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w^2+23w+24=0
a = 1; b = 23; c = +24;
Δ = b2-4ac
Δ = 232-4·1·24
Δ = 433
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{-(23)-\sqrt{433}}{2*1}=\frac{-23-\sqrt{433}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{433}}{2*1}=\frac{-23+\sqrt{433}}{2} $
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