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(3x^2)+(3x^2)-15=x
We move all terms to the left:
(3x^2)+(3x^2)-15-(x)=0
We add all the numbers together, and all the variables
6x^2-1x-15=0
a = 6; b = -1; c = -15;
Δ = b2-4ac
Δ = -12-4·6·(-15)
Δ = 361
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{361}=19$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-19}{2*6}=\frac{-18}{12} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+19}{2*6}=\frac{20}{12} =1+2/3 $
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