Y=-0.12x^2+2.8x

Solution for Y=-0.12x^2+2.8x equation:

=-0.12Y^2+2.8Y

We move all terms to the left:

-(-0.12Y^2+2.8Y)=0

We get rid of parentheses

0.12Y^2-2.8Y=0

a = 0.12; b = -2.8; c = 0;
Δ = b2-4ac
Δ = -2.82-4·0.12·0
Δ = 7.84
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{-(-2.8)-\sqrt{7.84}}{2*0.12}=\frac{2.8-\sqrt{7.84}}{0.24}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.8)+\sqrt{7.84}}{2*0.12}=\frac{2.8+\sqrt{7.84}}{0.24}
$