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w^2+22w-48=0
a = 1; b = 22; c = -48;
Δ = b2-4ac
Δ = 222-4·1·(-48)
Δ = 676
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{676}=26$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-26}{2*1}=\frac{-48}{2} =-24 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+26}{2*1}=\frac{4}{2} =2 $
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