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(n+3)(n-2)=54
We move all terms to the left:
(n+3)(n-2)-(54)=0
We multiply parentheses ..
(+n^2-2n+3n-6)-54=0
We get rid of parentheses
n^2-2n+3n-6-54=0
We add all the numbers together, and all the variables
n^2+n-60=0
a = 1; b = 1; c = -60;
Δ = b2-4ac
Δ = 12-4·1·(-60)
Δ = 241
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{241}}{2*1}=\frac{-1-\sqrt{241}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{241}}{2*1}=\frac{-1+\sqrt{241}}{2} $
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