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15v^2-14v+3=0
a = 15; b = -14; c = +3;
Δ = b2-4ac
Δ = -142-4·15·3
Δ = 16
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{16}=4$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4}{2*15}=\frac{10}{30} =1/3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4}{2*15}=\frac{18}{30} =3/5 $
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