40=1.2x+0.04x^2

Solution for 40=1.2x+0.04x^2 equation:

40=1.2x+0.04x^2

We move all terms to the left:

40-(1.2x+0.04x^2)=0

We get rid of parentheses

-0.04x^2-1.2x+40=0

a = -0.04; b = -1.2; c = +40;
Δ = b2-4ac
Δ = -1.22-4·(-0.04)·40
Δ = 7.84
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{7.84}}{2*-0.04}=\frac{1.2-\sqrt{7.84}}{-0.08}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.2)+\sqrt{7.84}}{2*-0.04}=\frac{1.2+\sqrt{7.84}}{-0.08}
$