(x+20)(x-4)=952

Solution for (x+20)(x-4)=952 equation:

(x+20)(x-4)=952

We move all terms to the left:

(x+20)(x-4)-(952)=0

We multiply parentheses ..

(+x^2-4x+20x-80)-952=0

We get rid of parentheses

x^2-4x+20x-80-952=0

We add all the numbers together, and all the variables

x^2+16x-1032=0

a = 1; b = 16; c = -1032;
Δ = b2-4ac
Δ = 162-4·1·(-1032)
Δ = 4384
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:

$\sqrt{\Delta}=\sqrt{4384}=\sqrt{16*274}=\sqrt{16}*\sqrt{274}=4\sqrt{274}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{274}}{2*1}=\frac{-16-4\sqrt{274}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{274}}{2*1}=\frac{-16+4\sqrt{274}}{2}
$