5x^2+20x=-16

Solution for 5x^2+20x=-16 equation:

5x^2+20x=-16

We move all terms to the left:

5x^2+20x-(-16)=0

We add all the numbers together, and all the variables

5x^2+20x+16=0

a = 5; b = 20; c = +16;
Δ = b2-4ac
Δ = 202-4·5·16
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{5}}{2*5}=\frac{-20-4\sqrt{5}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{5}}{2*5}=\frac{-20+4\sqrt{5}}{10}
$