-0.6x^2+18x-130=0

Solution for -0.6x^2+18x-130=0 equation:

-0.6x^2+18x-130=0

a = -0.6; b = 18; c = -130;
Δ = b2-4ac
Δ = 182-4·(-0.6)·(-130)
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{3}}{2*-0.6}=\frac{-18-2\sqrt{3}}{-1.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{3}}{2*-0.6}=\frac{-18+2\sqrt{3}}{-1.2}
$