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6w^2+23w-35=0
a = 6; b = 23; c = -35;
Δ = b2-4ac
Δ = 232-4·6·(-35)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1369}=37$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-37}{2*6}=\frac{-60}{12} =-5 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+37}{2*6}=\frac{14}{12} =1+1/6 $
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