12=-8u+6(u+5)u=

Solution for 12=-8u+6(u+5)u= equation:

12=-8u+6(u+5)u=

We move all terms to the left:

12-(-8u+6(u+5)u)=0


We calculate terms in parentheses: -(-8u+6(u+5)u), so:

-8u+6(u+5)u

We multiply parentheses

6u^2-8u+30u

We add all the numbers together, and all the variables

6u^2+22u

Back to the equation:

-(6u^2+22u)


We get rid of parentheses

-6u^2-22u+12=0

a = -6; b = -22; c = +12;
Δ = b2-4ac
Δ = -222-4·(-6)·12
Δ = 772
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{772}=\sqrt{4*193}=\sqrt{4}*\sqrt{193}=2\sqrt{193}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{193}}{2*-6}=\frac{22-2\sqrt{193}}{-12}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{193}}{2*-6}=\frac{22+2\sqrt{193}}{-12}
$