Y=-0.02x^2+0.4x+1.7

Solution for Y=-0.02x^2+0.4x+1.7 equation:

=-0.02Y^2+0.4Y+1.7

We move all terms to the left:

-(-0.02Y^2+0.4Y+1.7)=0

We get rid of parentheses

0.02Y^2-0.4Y-1.7=0

a = 0.02; b = -0.4; c = -1.7;
Δ = b2-4ac
Δ = -0.42-4·0.02·(-1.7)
Δ = 0.296
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{-(-0.4)-\sqrt{0.296}}{2*0.02}=\frac{0.4-\sqrt{0.296}}{0.04}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.4)+\sqrt{0.296}}{2*0.02}=\frac{0.4+\sqrt{0.296}}{0.04}
$