0=-2(x-3)(x-3)+4

Solution for 0=-2(x-3)(x-3)+4 equation:

0=-2(x-3)(x-3)+4

We move all terms to the left:

0-(-2(x-3)(x-3)+4)=0

We add all the numbers together, and all the variables

-(-2(x-3)(x-3)+4)=0

We multiply parentheses ..

-(-2(+x^2-3x-3x+9)+4)=0


We calculate terms in parentheses: -(-2(+x^2-3x-3x+9)+4), so:

-2(+x^2-3x-3x+9)+4

We multiply parentheses

-2x^2+6x+6x-18+4

We add all the numbers together, and all the variables

-2x^2+12x-14

Back to the equation:

-(-2x^2+12x-14)


We get rid of parentheses

2x^2-12x+14=0

a = 2; b = -12; c = +14;
Δ = b2-4ac
Δ = -122-4·2·14
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{2}}{2*2}=\frac{12-4\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{2}}{2*2}=\frac{12+4\sqrt{2}}{4}
$