Y=-0.12x^2+2.4x

Solution for Y=-0.12x^2+2.4x equation:

=-0.12Y^2+2.4Y

We move all terms to the left:

-(-0.12Y^2+2.4Y)=0

We get rid of parentheses

0.12Y^2-2.4Y=0

a = 0.12; b = -2.4; c = 0;
Δ = b2-4ac
Δ = -2.42-4·0.12·0
Δ = 5.76
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{-(-2.4)-\sqrt{5.76}}{2*0.12}=\frac{2.4-\sqrt{5.76}}{0.24}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.4)+\sqrt{5.76}}{2*0.12}=\frac{2.4+\sqrt{5.76}}{0.24}
$