What are the solutions to the equation (n+2)(n-4)=27 ?

The solutions to the equation (n+2)(n-4)=27 are n=7,n=-5

Find the zeros of (n+2)(n-4)=27

The zeros of (n+2)(n-4)=27 are n=7,n=-5

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(n + 2) (n - 4) = 27

n = 7, n = - 5

Solution steps

(n + 2) (n - 4) = 27

Expand

(n + 2) (n - 4) : n^{2} - 2 n - 8

n^{2} - 2 n - 8 = 27

Move

27

to the left side

n^{2} - 2 n - 35 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 35 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 35) = 12

n_{1, 2} = \frac{- ( - 2 ) \pm 12}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 2 ) + 12}{2 \cdot 1}, n_{2} = \frac{- ( - 2 ) - 12}{2 \cdot 1}

n = \frac{- ( - 2 ) + 12}{2 \cdot 1} : 7

n = \frac{- ( - 2 ) - 12}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

n = 7, n = - 5

6 - 3 y - y^{2} = 6

2 p^{2} - 11 = 61

3 x^{2} + 16 x - 2 = 0

3 v^{2} - 27 = 0

y, x^{2} - x y + \frac{1}{2} y^{2} = 3

What are the solutions to the equation (n+2)(n-4)=27 ?
The solutions to the equation (n+2)(n-4)=27 are n=7,n=-5
Find the zeros of (n+2)(n-4)=27
The zeros of (n+2)(n-4)=27 are n=7,n=-5