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w^2+11w+25=0
a = 1; b = 11; c = +25;
Δ = b2-4ac
Δ = 112-4·1·25
Δ = 21
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{21}}{2*1}=\frac{-11-\sqrt{21}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{21}}{2*1}=\frac{-11+\sqrt{21}}{2} $
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