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w^2-11w-50=0
a = 1; b = -11; c = -50;
Δ = b2-4ac
Δ = -112-4·1·(-50)
Δ = 321
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{321}}{2*1}=\frac{11-\sqrt{321}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{321}}{2*1}=\frac{11+\sqrt{321}}{2} $
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