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

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

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

We move all terms to the left:

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

We multiply parentheses

12u^2-3u-(2(u+2)-1)=0


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

2(u+2)-1

We multiply parentheses

2u+4-1

We add all the numbers together, and all the variables

2u+3

Back to the equation:

-(2u+3)


We get rid of parentheses

12u^2-3u-2u-3=0

We add all the numbers together, and all the variables

12u^2-5u-3=0

a = 12; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·12·(-3)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*12}=\frac{-8}{24}
=-1/3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*12}=\frac{18}{24}
=3/4
$