-0.03x^2+1.2x=3

Solution for -0.03x^2+1.2x=3 equation:

-0.03x^2+1.2x=3

We move all terms to the left:

-0.03x^2+1.2x-(3)=0

a = -0.03; b = 1.2; c = -3;
Δ = b2-4ac
Δ = 1.22-4·(-0.03)·(-3)
Δ = 1.08
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{1.08}}{2*-0.03}=\frac{-1.2-\sqrt{1.08}}{-0.06}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+\sqrt{1.08}}{2*-0.03}=\frac{-1.2+\sqrt{1.08}}{-0.06}
$