5a^2+20a=-4

Solution for 5a^2+20a=-4 equation:

5a^2+20a=-4

We move all terms to the left:

5a^2+20a-(-4)=0

We add all the numbers together, and all the variables

5a^2+20a+4=0

a = 5; b = 20; c = +4;
Δ = b2-4ac
Δ = 202-4·5·4
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{5}}{2*5}=\frac{-20-8\sqrt{5}}{10}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{5}}{2*5}=\frac{-20+8\sqrt{5}}{10}
$