5u^2+32u-64=0

Solution for 5u^2+32u-64=0 equation:

5u^2+32u-64=0

a = 5; b = 32; c = -64;
Δ = b2-4ac
Δ = 322-4·5·(-64)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-48}{2*5}=\frac{-80}{10}
=-8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+48}{2*5}=\frac{16}{10}
=1+3/5
$