x^2+20x=16

Solution for x^2+20x=16 equation:

x^2+20x=16

We move all terms to the left:

x^2+20x-(16)=0

a = 1; b = 20; c = -16;
Δ = b2-4ac
Δ = 202-4·1·(-16)
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{29}}{2*1}=\frac{-20-4\sqrt{29}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{29}}{2*1}=\frac{-20+4\sqrt{29}}{2}
$