9x^2-18+9=16x-12

Solution for 9x^2-18+9=16x-12 equation:

9x^2-18+9=16x-12

We move all terms to the left:

9x^2-18+9-(16x-12)=0

We add all the numbers together, and all the variables

9x^2-(16x-12)-9=0

We get rid of parentheses

9x^2-16x+12-9=0

We add all the numbers together, and all the variables

9x^2-16x+3=0

a = 9; b = -16; c = +3;
Δ = b2-4ac
Δ = -162-4·9·3
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{37}}{2*9}=\frac{16-2\sqrt{37}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{37}}{2*9}=\frac{16+2\sqrt{37}}{18}
$