4x^2+52x+40=0

Solution for 4x^2+52x+40=0 equation:

4x^2+52x+40=0

a = 4; b = 52; c = +40;
Δ = b2-4ac
Δ = 522-4·4·40
Δ = 2064
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:

$\sqrt{\Delta}=\sqrt{2064}=\sqrt{16*129}=\sqrt{16}*\sqrt{129}=4\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-4\sqrt{129}}{2*4}=\frac{-52-4\sqrt{129}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+4\sqrt{129}}{2*4}=\frac{-52+4\sqrt{129}}{8}
$