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6v^2+9v-10=0
a = 6; b = 9; c = -10;
Δ = b2-4ac
Δ = 92-4·6·(-10)
Δ = 321
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}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{321}}{2*6}=\frac{-9-\sqrt{321}}{12} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{321}}{2*6}=\frac{-9+\sqrt{321}}{12} $
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