-3x^2+48x-180=0

Solution for -3x^2+48x-180=0 equation:

-3x^2+48x-180=0

a = -3; b = 48; c = -180;
Δ = b2-4ac
Δ = 482-4·(-3)·(-180)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-12}{2*-3}=\frac{-60}{-6}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+12}{2*-3}=\frac{-36}{-6}
=+6
$