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x=(x-2)(3x+3)
We move all terms to the left:
x-((x-2)(3x+3))=0
We multiply parentheses ..
-((+3x^2+3x-6x-6))+x=0
We calculate terms in parentheses: -((+3x^2+3x-6x-6)), so:We add all the numbers together, and all the variables
(+3x^2+3x-6x-6)
We get rid of parentheses
3x^2+3x-6x-6
We add all the numbers together, and all the variables
3x^2-3x-6
Back to the equation:
-(3x^2-3x-6)
x-(3x^2-3x-6)=0
We get rid of parentheses
-3x^2+x+3x+6=0
We add all the numbers together, and all the variables
-3x^2+4x+6=0
a = -3; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·(-3)·6
Δ = 88
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{22}}{2*-3}=\frac{-4-2\sqrt{22}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{22}}{2*-3}=\frac{-4+2\sqrt{22}}{-6} $
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