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6x^2+x-35=0
a = 6; b = 1; c = -35;
Δ = b2-4ac
Δ = 12-4·6·(-35)
Δ = 841
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{841}=29$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-29}{2*6}=\frac{-30}{12} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+29}{2*6}=\frac{28}{12} =2+1/3 $
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