# Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Solve for x in the equation 2x^2+3x-7=x^2+5x+39

## This Post Has 10 Comments

1. Akira8889 says:

Hey there, hope I can help!

$\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}$
$2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)$

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
$x^2-2x-46=0$

Lets use the quadratic formula now
$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$
$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}$

$\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

Multiply the numbers 2 * 1 = 2
$\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ \sqrt{4+184} \ \textgreater \ \sqrt{188} \ \textgreater \ 2 + \sqrt{188}$
$\frac{2+\sqrt{188}}{2} \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ \sqrt{2^2\cdot \:47}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}$

$Factor\;2+2\sqrt{47} \ \textgreater \ Rewrite\;as\;1\cdot \:2+2\sqrt{47}$
$\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+\sqrt{47}\right) \ \textgreater \ \frac{2\left(1+\sqrt{47}\right)}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ 1+\sqrt{47}$

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

$\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

$\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ 2-\sqrt{188} \ \textgreater \ \frac{2-\sqrt{188}}{2}$

$\sqrt{188} = 2\sqrt{47} \ \textgreater \ \frac{2-2\sqrt{47}}{2}$

$2-2\sqrt{47} \ \textgreater \ 2\left(1-\sqrt{47}\right) \ \textgreater \ \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \ 1-\sqrt{47}$

Therefore our final solutions are
$x=1+\sqrt{47},\:x=1-\sqrt{47}$

Hope this helps!

2. markeledwards699 says:

$2x^2 - 3x - 4 = 0 \\ x= \frac{-b \pm \sqrt{ b^{2} -4ac} }{2a} \\ = \frac{-(-3) \pm \sqrt{ (-3)^{2} -4 \times2 \times (-4)} }{2 \times 2} \\ = \frac{3 \pm \sqrt{ 9 +32} }{4} \\ = \frac{3 \pm \sqrt{41} }{4} \\ = \frac{3 + \sqrt{41} }{4} \ or \ \frac{3 - \sqrt{41} }{4} \\ =2.35 \ or \ -0.85$

3. lagarde says:

D

hope this helps :o)

4. misaki2002 says:

Sorry I can't explain this one much but it is x = 1+- square root of 47

5. bsheepicornozj0gc says:

There is a combination of both answers  x=7.8556546,−5.8556546

6. areanna02 says:

$X=1+\sqrt{47}$

Step-by-step explanation:

Hope this helps

Mark me as brainiest

7. Baby010391 says:

x=1±√47

Step-by-step explanation:

it's up above.

$Solve for x in the equation 2x^2+3x-7=x^2+5x+39$

8. twalters88 says:

Subtract x^2 from both sides
x^2 + 3x - 7 = 5x + 39
Subtract 5x from both sides
x^2 - 2x - 7 = 39
x^2 - 2x = 46
Complete the square by adding (b/2)^2 to both sides, b = ( -2)
(-2/2) = -1, then square that (-1)^2 = 1
x^2 - 2x + 1 = 46 + 1
Simplify the expression by factoring
(x - 1)^2 = 47
Take square root on each side
x - 1 = (sqrt (47))
Solve for x
x = 1 + (sqrt (47))
Since 47 is prime, 47 cannot be broken down by the square root and this is the answer to your problem.

9. isabellemaine says:

x1= 1 + √47

x2= 1 - √47

I use a handheld calculator to solve the problem

10. jackieanguiano4758 says:

$x=1\pm\sqrt{47}$

Step-by-step explanation:

We have been given an equation $2x^2+3x-7=x^2+5x+39$. We are asked to find the solution for our given equation.

$2x^2+3x-7=x^2+5x+39$

$2x^2-x^2+3x-7=x^2-x^2+5x+39$

$x^2+3x-7=5x+39$

$x^2+3x-5x-7-39=5x-5x+39-39$

$x^2-2x-46=0$

Using quadratic formula, we will get:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-46)}}{2(1)}$

$x=\frac{2\pm\sqrt{4+184}}{2}$

$x=\frac{2\pm\sqrt{188}}{2}$

$x=\frac{2\pm2\sqrt{47}}{2}$

$x=1\pm\sqrt{47}$

Therefore, the solutions for our given equation are $x=1\pm\sqrt{47}$.