Y=-2x^2+3x+50

Solution for Y=-2x^2+3x+50 equation:

=-2Y^2+3Y+50

We move all terms to the left:

-(-2Y^2+3Y+50)=0

We get rid of parentheses

2Y^2-3Y-50=0

a = 2; b = -3; c = -50;
Δ = b2-4ac
Δ = -32-4·2·(-50)
Δ = 409
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{409}}{2*2}=\frac{3-\sqrt{409}}{4}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{409}}{2*2}=\frac{3+\sqrt{409}}{4}
$