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j^2-11j-10=0
a = 1; b = -11; c = -10;
Δ = b2-4ac
Δ = -112-4·1·(-10)
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{161}}{2*1}=\frac{11-\sqrt{161}}{2} $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{161}}{2*1}=\frac{11+\sqrt{161}}{2} $
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