0,12=-0,12x^2+6

Solution for 0,12=-0,12x^2+6 equation:

0.12=-0.12x^2+6

We move all terms to the left:

0.12-(-0.12x^2+6)=0

We get rid of parentheses

0.12x^2-6+0.12=0

We add all the numbers together, and all the variables

0.12x^2-5.88=0

a = 0.12; b = 0; c = -5.88;
Δ = b2-4ac
Δ = 02-4·0.12·(-5.88)
Δ = 2.8224
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{2.8224}}{2*0.12}=\frac{0-\sqrt{2.8224}}{0.24}
=-\frac{\sqrt{}}{0.24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{2.8224}}{2*0.12}=\frac{0+\sqrt{2.8224}}{0.24}
=\frac{\sqrt{}}{0.24}
$