18x^2+48x-96=0

Solution for 18x^2+48x-96=0 equation:

18x^2+48x-96=0

a = 18; b = 48; c = -96;
Δ = b2-4ac
Δ = 482-4·18·(-96)
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-96}{2*18}=\frac{-144}{36}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+96}{2*18}=\frac{48}{36}
=1+1/3
$