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3v^2+v-114=0
a = 3; b = 1; c = -114;
Δ = b2-4ac
Δ = 12-4·3·(-114)
Δ = 1369
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{1369}=37$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-37}{2*3}=\frac{-38}{6} =-6+1/3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+37}{2*3}=\frac{36}{6} =6 $
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