15x^2+132x+96=0

Solution for 15x^2+132x+96=0 equation:

15x^2+132x+96=0

a = 15; b = 132; c = +96;
Δ = b2-4ac
Δ = 1322-4·15·96
Δ = 11664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11664}=108$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(132)-108}{2*15}=\frac{-240}{30}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(132)+108}{2*15}=\frac{-24}{30}
=-4/5
$