Question

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

Answer 1

Subtract x^2 from both sides
x^2 + 3x - 7 = 5x + 39
Subtract 5x from both sides
x^2 - 2x - 7 = 39
Add 7 to both sides
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.

Answer 2

Answer:

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

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