3*(5-2x)=9x^2+9

Solution for 3*(5-2x)=9x^2+9 equation:

3(5-2x)=9x^2+9

We move all terms to the left:

3(5-2x)-(9x^2+9)=0

We add all the numbers together, and all the variables

3(-2x+5)-(9x^2+9)=0

We multiply parentheses

-6x-(9x^2+9)+15=0

We get rid of parentheses

-9x^2-6x-9+15=0

We add all the numbers together, and all the variables

-9x^2-6x+6=0

a = -9; b = -6; c = +6;
Δ = b2-4ac
Δ = -62-4·(-9)·6
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{7}}{2*-9}=\frac{6-6\sqrt{7}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{7}}{2*-9}=\frac{6+6\sqrt{7}}{-18}
$