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6x^2+18x=x-12
We move all terms to the left:
6x^2+18x-(x-12)=0
We get rid of parentheses
6x^2+18x-x+12=0
We add all the numbers together, and all the variables
6x^2+17x+12=0
a = 6; b = 17; c = +12;
Δ = b2-4ac
Δ = 172-4·6·12
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-1}{2*6}=\frac{-18}{12} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+1}{2*6}=\frac{-16}{12} =-1+1/3 $
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