0=3(x^2-6x+7)

Solution for 0=3(x^2-6x+7) equation:

0=3(x^2-6x+7)

We move all terms to the left:

0-(3(x^2-6x+7))=0

We add all the numbers together, and all the variables

-(3(x^2-6x+7))=0


We calculate terms in parentheses: -(3(x^2-6x+7)), so:

3(x^2-6x+7)

We multiply parentheses

3x^2-18x+21

Back to the equation:

-(3x^2-18x+21)


We get rid of parentheses

-3x^2+18x-21=0

a = -3; b = 18; c = -21;
Δ = b2-4ac
Δ = 182-4·(-3)·(-21)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{2}}{2*-3}=\frac{-18-6\sqrt{2}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{2}}{2*-3}=\frac{-18+6\sqrt{2}}{-6}
$