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v^2+11v-12=0
a = 1; b = 11; c = -12;
Δ = b2-4ac
Δ = 112-4·1·(-12)
Δ = 169
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{169}=13$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*1}=\frac{-24}{2} =-12 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*1}=\frac{2}{2} =1 $
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