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(n+5)(n-7)=13
We move all terms to the left:
(n+5)(n-7)-(13)=0
We multiply parentheses ..
(+n^2-7n+5n-35)-13=0
We get rid of parentheses
n^2-7n+5n-35-13=0
We add all the numbers together, and all the variables
n^2-2n-48=0
a = 1; b = -2; c = -48;
Δ = b2-4ac
Δ = -22-4·1·(-48)
Δ = 196
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{196}=14$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*1}=\frac{-12}{2} =-6 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*1}=\frac{16}{2} =8 $
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