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v2+5v-6=0
We add all the numbers together, and all the variables
v^2+5v-6=0
a = 1; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·1·(-6)
Δ = 49
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{49}=7$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*1}=\frac{-12}{2} =-6 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*1}=\frac{2}{2} =1 $
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