5(2x+2)=(x^2+2x)-23

Solution for 5(2x+2)=(x^2+2x)-23 equation:

5(2x+2)=(x^2+2x)-23

We move all terms to the left:

5(2x+2)-((x^2+2x)-23)=0

We multiply parentheses

10x-((x^2+2x)-23)+10=0


We calculate terms in parentheses: -((x^2+2x)-23), so:

(x^2+2x)-23

We get rid of parentheses

x^2+2x-23

Back to the equation:

-(x^2+2x-23)


We get rid of parentheses

-x^2+10x-2x+23+10=0

We add all the numbers together, and all the variables

-1x^2+8x+33=0

a = -1; b = 8; c = +33;
Δ = b2-4ac
Δ = 82-4·(-1)·33
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-14}{2*-1}=\frac{-22}{-2}
=+11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+14}{2*-1}=\frac{6}{-2}
=-3
$