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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-5u^2+28u+12=0

a = -5; b = 28; c = +12;
Δ = b2-4ac
Δ = 282-4·(-5)·12
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-32}{2*-5}=\frac{-60}{-10}
=+6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+32}{2*-5}=\frac{4}{-10}
=-2/5
$