12=0.03x^2

Solution for 12=0.03x^2 equation:

12=0.03x^2

We move all terms to the left:

12-(0.03x^2)=0

We get rid of parentheses

-0.03x^2+12=0

a = -0.03; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-0.03)·12
Δ = 1.44
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{-(0)-\sqrt{1.44}}{2*-0.03}=\frac{0-\sqrt{1.44}}{-0.06}
=-\frac{\sqrt{}}{-0.06}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{1.44}}{2*-0.03}=\frac{0+\sqrt{1.44}}{-0.06}
=\frac{\sqrt{}}{-0.06}
$