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15v^2+28v+12=0
a = 15; b = 28; c = +12;
Δ = b2-4ac
Δ = 282-4·15·12
Δ = 64
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{64}=8$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-8}{2*15}=\frac{-36}{30} =-1+1/5 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+8}{2*15}=\frac{-20}{30} =-2/3 $
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