4y^2+24y+20=0

Solution for 4y^2+24y+20=0 equation:

4y^2+24y+20=0

a = 4; b = 24; c = +20;
Δ = b2-4ac
Δ = 242-4·4·20
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16}{2*4}=\frac{-40}{8}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16}{2*4}=\frac{-8}{8}
=-1
$