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

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

-3d^2+48d-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:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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