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3x=x^2-6x-12
We move all terms to the left:
3x-(x^2-6x-12)=0
We get rid of parentheses
-x^2+3x+6x+12=0
We add all the numbers together, and all the variables
-1x^2+9x+12=0
a = -1; b = 9; c = +12;
Δ = b2-4ac
Δ = 92-4·(-1)·12
Δ = 129
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{129}}{2*-1}=\frac{-9-\sqrt{129}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{129}}{2*-1}=\frac{-9+\sqrt{129}}{-2} $
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