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n^2-22n-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:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{676}=26$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-26}{2*1}=\frac{-4}{2} =-2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+26}{2*1}=\frac{48}{2} =24 $
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