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6x^2+x-92=0
a = 6; b = 1; c = -92;
Δ = b2-4ac
Δ = 12-4·6·(-92)
Δ = 2209
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{2209}=47$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-47}{2*6}=\frac{-48}{12} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+47}{2*6}=\frac{46}{12} =3+5/6 $
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