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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((-1x^2+3x-2x+6)), so:

(-1x^2+3x-2x+6)

We get rid of parentheses

-1x^2+3x-2x+6

We add all the numbers together, and all the variables

-1x^2+x+6

Back to the equation:

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


We get rid of parentheses

x^2+1x^2-x-6-3=0

We add all the numbers together, and all the variables

2x^2-1x-9=0

a = 2; b = -1; c = -9;
Δ = b2-4ac
Δ = -12-4·2·(-9)
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{73}}{2*2}=\frac{1-\sqrt{73}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{73}}{2*2}=\frac{1+\sqrt{73}}{4}
$