Y=-5x^2+4x

Solution for Y=-5x^2+4x equation:

=-5Y^2+4Y

We move all terms to the left:

-(-5Y^2+4Y)=0

We get rid of parentheses

5Y^2-4Y=0

a = 5; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·5·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*5}=\frac{0}{10}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*5}=\frac{8}{10}
=4/5
$