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6x^2=12+x
We move all terms to the left:
6x^2-(12+x)=0
We add all the numbers together, and all the variables
6x^2-(x+12)=0
We get rid of parentheses
6x^2-x-12=0
We add all the numbers together, and all the variables
6x^2-1x-12=0
a = 6; b = -1; c = -12;
Δ = b2-4ac
Δ = -12-4·6·(-12)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*6}=\frac{-16}{12} =-1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*6}=\frac{18}{12} =1+1/2 $
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