4x^2+40x+28=0

Solution for 4x^2+40x+28=0 equation:

4x^2+40x+28=0

a = 4; b = 40; c = +28;
Δ = b2-4ac
Δ = 402-4·4·28
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-24\sqrt{2}}{2*4}=\frac{-40-24\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+24\sqrt{2}}{2*4}=\frac{-40+24\sqrt{2}}{8}
$