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2v^2+11v-13=0
a = 2; b = 11; c = -13;
Δ = b2-4ac
Δ = 112-4·2·(-13)
Δ = 225
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{225}=15$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-15}{2*2}=\frac{-26}{4} =-6+1/2 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+15}{2*2}=\frac{4}{4} =1 $
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