4x^2+28.65x+50.41=0

Solution for 4x^2+28.65x+50.41=0 equation:

4x^2+28.65x+50.41=0

a = 4; b = 28.65; c = +50.41;
Δ = b2-4ac
Δ = 28.652-4·4·50.41
Δ = 14.2625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28.65)-\sqrt{14.2625}}{2*4}=\frac{-28.65-\sqrt{14.2625}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28.65)+\sqrt{14.2625}}{2*4}=\frac{-28.65+\sqrt{14.2625}}{8}
$