5d^2+48d-180=0

Solution for 5d^2+48d-180=0 equation:

5d^2+48d-180=0

a = 5; b = 48; c = -180;
Δ = b2-4ac
Δ = 482-4·5·(-180)
Δ = 5904
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5904}=\sqrt{144*41}=\sqrt{144}*\sqrt{41}=12\sqrt{41}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-12\sqrt{41}}{2*5}=\frac{-48-12\sqrt{41}}{10}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+12\sqrt{41}}{2*5}=\frac{-48+12\sqrt{41}}{10}
$