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14v-5v^2+3=0
a = -5; b = 14; c = +3;
Δ = b2-4ac
Δ = 142-4·(-5)·3
Δ = 256
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{256}=16$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-16}{2*-5}=\frac{-30}{-10} =+3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+16}{2*-5}=\frac{2}{-10} =-1/5 $
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