18x^2+96x-72=0

Solution for 18x^2+96x-72=0 equation:

18x^2+96x-72=0

a = 18; b = 96; c = -72;
Δ = b2-4ac
Δ = 962-4·18·(-72)
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-120}{2*18}=\frac{-216}{36}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+120}{2*18}=\frac{24}{36}
=2/3
$