-2d^2+100d-1200=0

Solution for -2d^2+100d-1200=0 equation:

-2d^2+100d-1200=0

a = -2; b = 100; c = -1200;
Δ = b2-4ac
Δ = 1002-4·(-2)·(-1200)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-20}{2*-2}=\frac{-120}{-4}
=+30
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+20}{2*-2}=\frac{-80}{-4}
=+20
$