-3x^2+40x-100=0

Solution for -3x^2+40x-100=0 equation:

-3x^2+40x-100=0

a = -3; b = 40; c = -100;
Δ = b2-4ac
Δ = 402-4·(-3)·(-100)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20}{2*-3}=\frac{-60}{-6}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20}{2*-3}=\frac{-20}{-6}
=3+1/3
$