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

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### 3 Answers

Given the equation:

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

To solve, first we will open the brackets:

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

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

Now we will multiply by 2:

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

Now we will subtract 5 from both sides:

==> 2n^2 - 2n -12 -5 = 0

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

Now we will use the formula to find the roots of the quadratic equation.

==> n1= ( 2 + sqrt(4+4*2*17) ]/ 2*2

= ( 2+ sqrt140) / 4

= 2+ 2sqrt(35) /4 = (1+ sqrt35)/2

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

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

To find n from (n+2)(n-3)=5/2, we first create quadratic equation

(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 - 12 - 5 = 0

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

Now n1 = [-b + sqrt ( b^2 - 4ac)]/2a

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

=> n1 = 1/2 + (sqrt 35)/2

n2 = 1/2- (sqrt 35)/2

**Therefore n is equal to 1/2 + (sqrt 35)/2 and 1/2 - (sqrt 35)/2.**

We'll multiply by 2 both sides:

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

We'll simplify and we'll get:

(n + 2)(2n - 6) = 5

We'll remove the brackets using the distributive law:

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**