Y=-0.2x^2-3x+20

Solution for Y=-0.2x^2-3x+20 equation:

=-0.2Y^2-3Y+20

We move all terms to the left:

-(-0.2Y^2-3Y+20)=0

We get rid of parentheses

0.2Y^2+3Y-20=0

a = 0.2; b = 3; c = -20;
Δ = b2-4ac
Δ = 32-4·0.2·(-20)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*0.2}=\frac{-8}{0.4}
=-20
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*0.2}=\frac{2}{0.4}
=5
$