(x+2)*(3+x)=(3-x)2+5x

Solution for (x+2)*(3+x)=(3-x)2+5x equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((-1x+3)2+5x), so:

(-1x+3)2+5x

We add all the numbers together, and all the variables

5x+(-1x+3)2

We multiply parentheses

5x-2x+6

We add all the numbers together, and all the variables

3x+6

Back to the equation:

-(3x+6)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+2x=0

a = 1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·1·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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