What are the solutions to the equation (n-6)(n-2)=-9n+32 ?

The solutions to the equation (n-6)(n-2)=-9n+32 are n=4,n=-5

Find the zeros of (n-6)(n-2)=-9n+32

The zeros of (n-6)(n-2)=-9n+32 are n=4,n=-5

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(n - 6) (n - 2) = - 9 n + 32

n = 4, n = - 5

Solution steps

(n - 6) (n - 2) = - 9 n + 32

Expand

(n - 6) (n - 2) : n^{2} - 8 n + 12

n^{2} - 8 n + 12 = - 9 n + 32

Move

32

to the left side

n^{2} - 8 n - 20 = - 9 n

Move

9 n

to the left side

n^{2} + n - 20 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 20) = 9

n_{1, 2} = \frac{- 1 \pm 9}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 1 + 9}{2 \cdot 1}, n_{2} = \frac{- 1 - 9}{2 \cdot 1}

n = \frac{- 1 + 9}{2 \cdot 1} : 4

n = \frac{- 1 - 9}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

n = 4, n = - 5

3 x^{2} - 16 x + 24 = 0

p, p^{1} \frac{v ^{1}}{t ^{1}} = p^{2} \frac{v ^{2}}{t ^{2}}

(x + 1) - x (x - 3) = 6

x^{2} + 20 x + 99 = 0

3 x^{2} = 8 x

What are the solutions to the equation (n-6)(n-2)=-9n+32 ?
The solutions to the equation (n-6)(n-2)=-9n+32 are n=4,n=-5
Find the zeros of (n-6)(n-2)=-9n+32
The zeros of (n-6)(n-2)=-9n+32 are n=4,n=-5