5d2+25d-110=0

Solution for 5d2+25d-110=0 equation:

5d^2+25d-110=0

a = 5; b = 25; c = -110;
Δ = b2-4ac
Δ = 252-4·5·(-110)
Δ = 2825
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:

$\sqrt{\Delta}=\sqrt{2825}=\sqrt{25*113}=\sqrt{25}*\sqrt{113}=5\sqrt{113}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{113}}{2*5}=\frac{-25-5\sqrt{113}}{10}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{113}}{2*5}=\frac{-25+5\sqrt{113}}{10}
$