6(x+1)+5=3x(2+x)

Solution for 6(x+1)+5=3x(2+x) equation:

6(x+1)+5=3x(2+x)

We move all terms to the left:

6(x+1)+5-(3x(2+x))=0

We add all the numbers together, and all the variables

6(x+1)-(3x(x+2))+5=0

We multiply parentheses

6x-(3x(x+2))+6+5=0


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

3x(x+2)

We multiply parentheses

3x^2+6x

Back to the equation:

-(3x^2+6x)


We add all the numbers together, and all the variables

6x-(3x^2+6x)+11=0

We get rid of parentheses

-3x^2+6x-6x+11=0

We add all the numbers together, and all the variables

-3x^2+11=0

a = -3; b = 0; c = +11;
Δ = b2-4ac
Δ = 02-4·(-3)·11
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{33}}{2*-3}=\frac{0-2\sqrt{33}}{-6}
=-\frac{2\sqrt{33}}{-6}
=-\frac{\sqrt{33}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{33}}{2*-3}=\frac{0+2\sqrt{33}}{-6}
=\frac{2\sqrt{33}}{-6}
=\frac{\sqrt{33}}{-3}
$