20x+x^2=256

Solution for 20x+x^2=256 equation:

20x+x^2=256

We move all terms to the left:

20x+x^2-(256)=0

a = 1; b = 20; c = -256;
Δ = b2-4ac
Δ = 202-4·1·(-256)
Δ = 1424
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:

$\sqrt{\Delta}=\sqrt{1424}=\sqrt{16*89}=\sqrt{16}*\sqrt{89}=4\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{89}}{2*1}=\frac{-20-4\sqrt{89}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{89}}{2*1}=\frac{-20+4\sqrt{89}}{2}
$