-3x^2+40x-35=0

Solution for -3x^2+40x-35=0 equation:

-3x^2+40x-35=0

a = -3; b = 40; c = -35;
Δ = b2-4ac
Δ = 402-4·(-3)·(-35)
Δ = 1180
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:

$\sqrt{\Delta}=\sqrt{1180}=\sqrt{4*295}=\sqrt{4}*\sqrt{295}=2\sqrt{295}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{295}}{2*-3}=\frac{-40-2\sqrt{295}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{295}}{2*-3}=\frac{-40+2\sqrt{295}}{-6}
$