4y^2+32y+28=0

Solution for 4y^2+32y+28=0 equation:

4y^2+32y+28=0

a = 4; b = 32; c = +28;
Δ = b2-4ac
Δ = 322-4·4·28
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-24}{2*4}=\frac{-56}{8}
=-7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+24}{2*4}=\frac{-8}{8}
=-1
$