-5x2+40x-63=0

Solution for -5x2+40x-63=0 equation:

-5x^2+40x-63=0

a = -5; b = 40; c = -63;
Δ = b2-4ac
Δ = 402-4·(-5)·(-63)
Δ = 340
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:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{85}}{2*-5}=\frac{-40-2\sqrt{85}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{85}}{2*-5}=\frac{-40+2\sqrt{85}}{-10}
$