9x^2-18x+9=16

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

9x^2-18x+9=16

We move all terms to the left:

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

We add all the numbers together, and all the variables

9x^2-18x-7=0

a = 9; b = -18; c = -7;
Δ = b2-4ac
Δ = -182-4·9·(-7)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-24}{2*9}=\frac{-6}{18}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+24}{2*9}=\frac{42}{18}
=2+1/3
$