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12=x^2+3x-6
We move all terms to the left:
12-(x^2+3x-6)=0
We get rid of parentheses
-x^2-3x+6+12=0
We add all the numbers together, and all the variables
-1x^2-3x+18=0
a = -1; b = -3; c = +18;
Δ = b2-4ac
Δ = -32-4·(-1)·18
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*-1}=\frac{-6}{-2} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*-1}=\frac{12}{-2} =-6 $
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