2(u+1)+7=3(u-2)2u

Solution for 2(u+1)+7=3(u-2)2u equation:

2(u+1)+7=3(u-2)2u

We move all terms to the left:

2(u+1)+7-(3(u-2)2u)=0

We multiply parentheses

2u-(3(u-2)2u)+2+7=0


We calculate terms in parentheses: -(3(u-2)2u), so:

3(u-2)2u

We multiply parentheses

6u^2-12u

Back to the equation:

-(6u^2-12u)


We add all the numbers together, and all the variables

2u-(6u^2-12u)+9=0

We get rid of parentheses

-6u^2+2u+12u+9=0

We add all the numbers together, and all the variables

-6u^2+14u+9=0

a = -6; b = 14; c = +9;
Δ = b2-4ac
Δ = 142-4·(-6)·9
Δ = 412
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:

$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{103}}{2*-6}=\frac{-14-2\sqrt{103}}{-12}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{103}}{2*-6}=\frac{-14+2\sqrt{103}}{-12}
$