-4y^2+50y+200=0

Solution for -4y^2+50y+200=0 equation:

-4y^2+50y+200=0

a = -4; b = 50; c = +200;
Δ = b2-4ac
Δ = 502-4·(-4)·200
Δ = 5700
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5700}=\sqrt{100*57}=\sqrt{100}*\sqrt{57}=10\sqrt{57}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{57}}{2*-4}=\frac{-50-10\sqrt{57}}{-8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{57}}{2*-4}=\frac{-50+10\sqrt{57}}{-8}
$