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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


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

2(x^2+3x-3)

We multiply parentheses

2x^2+6x-6

Back to the equation:

-(2x^2+6x-6)


We get rid of parentheses

-2x^2-6x+6=0

a = -2; b = -6; c = +6;
Δ = b2-4ac
Δ = -62-4·(-2)·6
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{21}}{2*-2}=\frac{6-2\sqrt{21}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{21}}{2*-2}=\frac{6+2\sqrt{21}}{-4}
$