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w^2+5w-33=0
a = 1; b = 5; c = -33;
Δ = b2-4ac
Δ = 52-4·1·(-33)
Δ = 157
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{-(5)-\sqrt{157}}{2*1}=\frac{-5-\sqrt{157}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{157}}{2*1}=\frac{-5+\sqrt{157}}{2} $
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