2(u+1)4u=3(2u-1)+8

Solution for 2(u+1)4u=3(2u-1)+8 equation:

2(u+1)4u=3(2u-1)+8

We move all terms to the left:

2(u+1)4u-(3(2u-1)+8)=0

We multiply parentheses

8u^2+8u-(3(2u-1)+8)=0


We calculate terms in parentheses: -(3(2u-1)+8), so:

3(2u-1)+8

We multiply parentheses

6u-3+8

We add all the numbers together, and all the variables

6u+5

Back to the equation:

-(6u+5)


We get rid of parentheses

8u^2+8u-6u-5=0

We add all the numbers together, and all the variables

8u^2+2u-5=0

a = 8; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·8·(-5)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{41}}{2*8}=\frac{-2-2\sqrt{41}}{16}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{41}}{2*8}=\frac{-2+2\sqrt{41}}{16}
$