4x^2+50x+96=0

Solution for 4x^2+50x+96=0 equation:

4x^2+50x+96=0

a = 4; b = 50; c = +96;
Δ = b2-4ac
Δ = 502-4·4·96
Δ = 964
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:

$\sqrt{\Delta}=\sqrt{964}=\sqrt{4*241}=\sqrt{4}*\sqrt{241}=2\sqrt{241}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{241}}{2*4}=\frac{-50-2\sqrt{241}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{241}}{2*4}=\frac{-50+2\sqrt{241}}{8}
$