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n^2-13n+56=8+n
We move all terms to the left:
n^2-13n+56-(8+n)=0
We add all the numbers together, and all the variables
n^2-13n-(n+8)+56=0
We get rid of parentheses
n^2-13n-n-8+56=0
We add all the numbers together, and all the variables
n^2-14n+48=0
a = 1; b = -14; c = +48;
Δ = b2-4ac
Δ = -142-4·1·48
Δ = 4
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{4}=2$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2}{2*1}=\frac{12}{2} =6 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2}{2*1}=\frac{16}{2} =8 $
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