-5u^2+20u+32=0

Solution for -5u^2+20u+32=0 equation:

-5u^2+20u+32=0

a = -5; b = 20; c = +32;
Δ = b2-4ac
Δ = 202-4·(-5)·32
Δ = 1040
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1040}=\sqrt{16*65}=\sqrt{16}*\sqrt{65}=4\sqrt{65}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{65}}{2*-5}=\frac{-20-4\sqrt{65}}{-10}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{65}}{2*-5}=\frac{-20+4\sqrt{65}}{-10}
$