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n2+26n+48=0
We add all the numbers together, and all the variables
n^2+26n+48=0
a = 1; b = 26; c = +48;
Δ = b2-4ac
Δ = 262-4·1·48
Δ = 484
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{484}=22$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-22}{2*1}=\frac{-48}{2} =-24 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+22}{2*1}=\frac{-4}{2} =-2 $
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