(4u+5)(6+-1u)=0

Solution for (4u+5)(6+-1u)=0 equation:

(4u+5)(6+-1u)=0

We add all the numbers together, and all the variables

(4u+5)(-1u)=0

We multiply parentheses ..

(-4u^2-5u)=0

We get rid of parentheses

-4u^2-5u=0

a = -4; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-4)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-4}=\frac{0}{-8}
=0
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-4}=\frac{10}{-8}
=-1+1/4
$