# What is n if (n+2)(n-3)=5/2? What is n if (n+2)(n-3)=5/2?

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We need to find n if (n+2)(n-3)=5/2

(n+2)(n-3)=5/2

=> n^2 + 2n - 3n - 6 = 5/2

=> n^2 - n - 6 = 5/2

=> 2n^2 - 2n - 12 = 5

=> 2n^2 - 2n - 17 = 0

n1 = 2/4 + sqrt(4 + 136)/4

=> 1/2 + sqrt 140 / 4

=> 0.5 + sqrt 35 / 2

n2 = 0.5 - sqrt 35 / 2

**The value of n is n = 0.5 + sqrt 35 / 2 and n = 0.5 - sqrt 35 / 2**

We'll multiply by 2 to the left:

2(n+2)(n-3)=5

We'll remove the brackets using FOIL method:

2n^2 - 6n + 4n - 12 = 5

We'll combine like terms and w'ell subtract 5 both sides:

2n^2 - 2n - 17 = 0

We'll apply quadratic formula:

n1 = [2 + sqrt(4 + 136)]/4

n1 = [2 + sqrt (140)]/4

**n1 = (1 + sqrt35)/2**

**n2 = (1 - sqrt35)/2**