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12v^2-5v-2=0
a = 12; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·12·(-2)
Δ = 121
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{121}=11$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*12}=\frac{-6}{24} =-1/4 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*12}=\frac{16}{24} =2/3 $
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