(3x^2-9)=(2x+9)

Solution for (3x^2-9)=(2x+9) equation:

(3x^2-9)=(2x+9)

We move all terms to the left:

(3x^2-9)-((2x+9))=0

We get rid of parentheses

3x^2-((2x+9))-9=0


We calculate terms in parentheses: -((2x+9)), so:

(2x+9)

We get rid of parentheses

2x+9

Back to the equation:

-(2x+9)


We get rid of parentheses

3x^2-2x-9-9=0

We add all the numbers together, and all the variables

3x^2-2x-18=0

a = 3; b = -2; c = -18;
Δ = b2-4ac
Δ = -22-4·3·(-18)
Δ = 220
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:

$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{55}}{2*3}=\frac{2-2\sqrt{55}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{55}}{2*3}=\frac{2+2\sqrt{55}}{6}
$