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9v^2+v-3=0
a = 9; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·9·(-3)
Δ = 109
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{-(1)-\sqrt{109}}{2*9}=\frac{-1-\sqrt{109}}{18} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{109}}{2*9}=\frac{-1+\sqrt{109}}{18} $
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