9x+27+9x=x^2+3x

Solution for 9x+27+9x=x^2+3x equation:

9x+27+9x=x^2+3x

We move all terms to the left:

9x+27+9x-(x^2+3x)=0

We add all the numbers together, and all the variables

18x-(x^2+3x)+27=0

We get rid of parentheses

-x^2+18x-3x+27=0

We add all the numbers together, and all the variables

-1x^2+15x+27=0

a = -1; b = 15; c = +27;
Δ = b2-4ac
Δ = 152-4·(-1)·27
Δ = 333
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:

$\sqrt{\Delta}=\sqrt{333}=\sqrt{9*37}=\sqrt{9}*\sqrt{37}=3\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{37}}{2*-1}=\frac{-15-3\sqrt{37}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{37}}{2*-1}=\frac{-15+3\sqrt{37}}{-2}
$