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

Solution for (x^2)-9x+9=0 equation:

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

a = 1; b = -9; c = +9;
Δ = b2-4ac
Δ = -92-4·1·9
Δ = 45
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:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{5}}{2*1}=\frac{9-3\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{5}}{2*1}=\frac{9+3\sqrt{5}}{2}
$