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2(x+1)=3x(x+2)
We move all terms to the left:
2(x+1)-(3x(x+2))=0
We multiply parentheses
2x-(3x(x+2))+2=0
We calculate terms in parentheses: -(3x(x+2)), so:We get rid of parentheses
3x(x+2)
We multiply parentheses
3x^2+6x
Back to the equation:
-(3x^2+6x)
-3x^2+2x-6x+2=0
We add all the numbers together, and all the variables
-3x^2-4x+2=0
a = -3; b = -4; c = +2;
Δ = b2-4ac
Δ = -42-4·(-3)·2
Δ = 40
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{10}}{2*-3}=\frac{4-2\sqrt{10}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{10}}{2*-3}=\frac{4+2\sqrt{10}}{-6} $
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