-0.3x^2+36x-1040=0

Solution for -0.3x^2+36x-1040=0 equation:

-0.3x^2+36x-1040=0

a = -0.3; b = 36; c = -1040;
Δ = b2-4ac
Δ = 362-4·(-0.3)·(-1040)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{3}}{2*-0.3}=\frac{-36-4\sqrt{3}}{-0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{3}}{2*-0.3}=\frac{-36+4\sqrt{3}}{-0.6}
$