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-3(x+4)=-3x(x+2)-6
We move all terms to the left:
-3(x+4)-(-3x(x+2)-6)=0
We multiply parentheses
-3x-(-3x(x+2)-6)-12=0
We calculate terms in parentheses: -(-3x(x+2)-6), so:We get rid of parentheses
-3x(x+2)-6
We multiply parentheses
-3x^2-6x-6
Back to the equation:
-(-3x^2-6x-6)
3x^2+6x-3x+6-12=0
We add all the numbers together, and all the variables
3x^2+3x-6=0
a = 3; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·3·(-6)
Δ = 81
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*3}=\frac{-12}{6} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*3}=\frac{6}{6} =1 $
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