4d^2+20d-25=0

Solution for 4d^2+20d-25=0 equation:

4d^2+20d-25=0

a = 4; b = 20; c = -25;
Δ = b2-4ac
Δ = 202-4·4·(-25)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{2}}{2*4}=\frac{-20-20\sqrt{2}}{8}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{2}}{2*4}=\frac{-20+20\sqrt{2}}{8}
$