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4v^2-13v-1=0
a = 4; b = -13; c = -1;
Δ = b2-4ac
Δ = -132-4·4·(-1)
Δ = 185
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{-(-13)-\sqrt{185}}{2*4}=\frac{13-\sqrt{185}}{8} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{185}}{2*4}=\frac{13+\sqrt{185}}{8} $
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