-5x^2+40x-45=0

Solution for -5x^2+40x-45=0 equation:

-5x^2+40x-45=0

a = -5; b = 40; c = -45;
Δ = b2-4ac
Δ = 402-4·(-5)·(-45)
Δ = 700
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:

$\sqrt{\Delta}=\sqrt{700}=\sqrt{100*7}=\sqrt{100}*\sqrt{7}=10\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{7}}{2*-5}=\frac{-40-10\sqrt{7}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{7}}{2*-5}=\frac{-40+10\sqrt{7}}{-10}
$