-0.12x^2+6=0

Solution for -0.12x^2+6=0 equation:

-0.12x^2+6=0

a = -0.12; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-0.12)·6
Δ = 2.88
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{2.88}}{2*-0.12}=\frac{0-\sqrt{2.88}}{-0.24}
=-\frac{\sqrt{}}{-0.24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{2.88}}{2*-0.12}=\frac{0+\sqrt{2.88}}{-0.24}
=\frac{\sqrt{}}{-0.24}
$