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6v^2+41v-7=0
a = 6; b = 41; c = -7;
Δ = b2-4ac
Δ = 412-4·6·(-7)
Δ = 1849
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{1849}=43$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-43}{2*6}=\frac{-84}{12} =-7 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+43}{2*6}=\frac{2}{12} =1/6 $
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