4x-2=(x^2)-4x-38

Solution for 4x-2=(x^2)-4x-38 equation:

4x-2=(x^2)-4x-38

We move all terms to the left:

4x-2-((x^2)-4x-38)=0

We get rid of parentheses

-x^2+4x+4x+38-2=0

We add all the numbers together, and all the variables

-1x^2+8x+36=0

a = -1; b = 8; c = +36;
Δ = b2-4ac
Δ = 82-4·(-1)·36
Δ = 208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{13}}{2*-1}=\frac{-8-4\sqrt{13}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{13}}{2*-1}=\frac{-8+4\sqrt{13}}{-2}
$