(x+20)(x-4)=1904

Solution for (x+20)(x-4)=1904 equation:

(x+20)(x-4)=1904

We move all terms to the left:

(x+20)(x-4)-(1904)=0

We multiply parentheses ..

(+x^2-4x+20x-80)-1904=0

We get rid of parentheses

x^2-4x+20x-80-1904=0

We add all the numbers together, and all the variables

x^2+16x-1984=0

a = 1; b = 16; c = -1984;
Δ = b2-4ac
Δ = 162-4·1·(-1984)
Δ = 8192
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:

$\sqrt{\Delta}=\sqrt{8192}=\sqrt{4096*2}=\sqrt{4096}*\sqrt{2}=64\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-64\sqrt{2}}{2*1}=\frac{-16-64\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+64\sqrt{2}}{2*1}=\frac{-16+64\sqrt{2}}{2}
$