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9v^2+28v+3=0
a = 9; b = 28; c = +3;
Δ = b2-4ac
Δ = 282-4·9·3
Δ = 676
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{676}=26$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-26}{2*9}=\frac{-54}{18} =-3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+26}{2*9}=\frac{-2}{18} =-1/9 $
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