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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

9x^2-34x+21=0

a = 9; b = -34; c = +21;
Δ = b2-4ac
Δ = -342-4·9·21
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-20}{2*9}=\frac{14}{18}
=7/9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+20}{2*9}=\frac{54}{18}
=3
$