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v^2-6v-55=0
a = 1; b = -6; c = -55;
Δ = b2-4ac
Δ = -62-4·1·(-55)
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$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-16}{2*1}=\frac{-10}{2} =-5 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+16}{2*1}=\frac{22}{2} =11 $
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