(6-u)(2u+5)=0

Solution for (6-u)(2u+5)=0 equation:

(6-u)(2u+5)=0

We add all the numbers together, and all the variables

(-1u+6)(2u+5)=0

We multiply parentheses ..

(-2u^2-5u+12u+30)=0

We get rid of parentheses

-2u^2-5u+12u+30=0

We add all the numbers together, and all the variables

-2u^2+7u+30=0

a = -2; b = 7; c = +30;
Δ = b2-4ac
Δ = 72-4·(-2)·30
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*-2}=\frac{-24}{-4}
=+6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*-2}=\frac{10}{-4}
=-2+1/2
$