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0=12x^2+6x
We move all terms to the left:
0-(12x^2+6x)=0
We add all the numbers together, and all the variables
-(12x^2+6x)=0
We get rid of parentheses
-12x^2-6x=0
a = -12; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·(-12)·0
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*-12}=\frac{0}{-24} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*-12}=\frac{12}{-24} =-1/2 $
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