10=0.02x^2+1.2x

Solution for 10=0.02x^2+1.2x equation:

10=0.02x^2+1.2x

We move all terms to the left:

10-(0.02x^2+1.2x)=0

We get rid of parentheses

-0.02x^2-1.2x+10=0

a = -0.02; b = -1.2; c = +10;
Δ = b2-4ac
Δ = -1.22-4·(-0.02)·10
Δ = 2.24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1.2)-\sqrt{2.24}}{2*-0.02}=\frac{1.2-\sqrt{2.24}}{-0.04}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.2)+\sqrt{2.24}}{2*-0.02}=\frac{1.2+\sqrt{2.24}}{-0.04}
$