(u+2)(3u-1)+1=(6-u)(u+1)

Solution for (u+2)(3u-1)+1=(6-u)(u+1) equation:

(u+2)(3u-1)+1=(6-u)(u+1)

We move all terms to the left:

(u+2)(3u-1)+1-((6-u)(u+1))=0

We add all the numbers together, and all the variables

(u+2)(3u-1)-((-1u+6)(u+1))+1=0

We multiply parentheses ..

(+3u^2-1u+6u-2)-((-1u+6)(u+1))+1=0


We calculate terms in parentheses: -((-1u+6)(u+1)), so:

(-1u+6)(u+1)

We multiply parentheses ..

(-1u^2-1u+6u+6)

We get rid of parentheses

-1u^2-1u+6u+6

We add all the numbers together, and all the variables

-1u^2+5u+6

Back to the equation:

-(-1u^2+5u+6)


We get rid of parentheses

3u^2+1u^2-1u+6u-5u-2-6+1=0

We add all the numbers together, and all the variables

4u^2-7=0

a = 4; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·4·(-7)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{7}}{2*4}=\frac{0-4\sqrt{7}}{8}
=-\frac{4\sqrt{7}}{8}
=-\frac{\sqrt{7}}{2}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{7}}{2*4}=\frac{0+4\sqrt{7}}{8}
=\frac{4\sqrt{7}}{8}
=\frac{\sqrt{7}}{2}
$