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7v^2+11v-6=0
a = 7; b = 11; c = -6;
Δ = b2-4ac
Δ = 112-4·7·(-6)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{289}=17$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-17}{2*7}=\frac{-28}{14} =-2 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+17}{2*7}=\frac{6}{14} =3/7 $
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