(8u+5)(4u-5)=0

Solution for (8u+5)(4u-5)=0 equation:

(8u+5)(4u-5)=0

We multiply parentheses ..

(+32u^2-40u+20u-25)=0

We get rid of parentheses

32u^2-40u+20u-25=0

We add all the numbers together, and all the variables

32u^2-20u-25=0

a = 32; b = -20; c = -25;
Δ = b2-4ac
Δ = -202-4·32·(-25)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-60}{2*32}=\frac{-40}{64}
=-5/8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+60}{2*32}=\frac{80}{64}
=1+1/4
$