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3x^2-3x-34=2
We move all terms to the left:
3x^2-3x-34-(2)=0
We add all the numbers together, and all the variables
3x^2-3x-36=0
a = 3; b = -3; c = -36;
Δ = b2-4ac
Δ = -32-4·3·(-36)
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-21}{2*3}=\frac{-18}{6} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+21}{2*3}=\frac{24}{6} =4 $
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