3x^2+2x=18x+9

Solution for 3x^2+2x=18x+9 equation:

3x^2+2x=18x+9

We move all terms to the left:

3x^2+2x-(18x+9)=0

We get rid of parentheses

3x^2+2x-18x-9=0

We add all the numbers together, and all the variables

3x^2-16x-9=0

a = 3; b = -16; c = -9;
Δ = b2-4ac
Δ = -162-4·3·(-9)
Δ = 364
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:

$\sqrt{\Delta}=\sqrt{364}=\sqrt{4*91}=\sqrt{4}*\sqrt{91}=2\sqrt{91}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{91}}{2*3}=\frac{16-2\sqrt{91}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{91}}{2*3}=\frac{16+2\sqrt{91}}{6}
$