5(x+x+2+x+4)=66+x^2+2x

Solution for 5(x+x+2+x+4)=66+x^2+2x equation:

5(x+x+2+x+4)=66+x^2+2x

We move all terms to the left:

5(x+x+2+x+4)-(66+x^2+2x)=0

We add all the numbers together, and all the variables

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

We multiply parentheses

-(66+x^2+2x)+15x+30=0

We get rid of parentheses

-x^2-2x+15x-66+30=0

We add all the numbers together, and all the variables

-1x^2+13x-36=0

a = -1; b = 13; c = -36;
Δ = b2-4ac
Δ = 132-4·(-1)·(-36)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*-1}=\frac{-18}{-2}
=+9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*-1}=\frac{-8}{-2}
=+4
$