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3x^2+x=68
We move all terms to the left:
3x^2+x-(68)=0
a = 3; b = 1; c = -68;
Δ = b2-4ac
Δ = 12-4·3·(-68)
Δ = 817
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{-(1)-\sqrt{817}}{2*3}=\frac{-1-\sqrt{817}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{817}}{2*3}=\frac{-1+\sqrt{817}}{6} $
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