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2x(3x-1)=x+8
We move all terms to the left:
2x(3x-1)-(x+8)=0
We multiply parentheses
6x^2-2x-(x+8)=0
We get rid of parentheses
6x^2-2x-x-8=0
We add all the numbers together, and all the variables
6x^2-3x-8=0
a = 6; b = -3; c = -8;
Δ = b2-4ac
Δ = -32-4·6·(-8)
Δ = 201
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{-(-3)-\sqrt{201}}{2*6}=\frac{3-\sqrt{201}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{201}}{2*6}=\frac{3+\sqrt{201}}{12} $
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