0=-(x-3)(x-3)+9

Solution for 0=-(x-3)(x-3)+9 equation:

0=-(x-3)(x-3)+9

We move all terms to the left:

0-(-(x-3)(x-3)+9)=0

We add all the numbers together, and all the variables

-(-(x-3)(x-3)+9)=0

We multiply parentheses ..

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


We calculate terms in parentheses: -(-(+x^2-3x-3x+9)+9), so:

-(+x^2-3x-3x+9)+9

We get rid of parentheses

-x^2+3x+3x-9+9

We add all the numbers together, and all the variables

-1x^2+6x

Back to the equation:

-(-1x^2+6x)


We get rid of parentheses

1x^2-6x=0

We add all the numbers together, and all the variables

x^2-6x=0

a = 1; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·1·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*1}=\frac{12}{2}
=6
$