What are the solutions to the equation n(4+2n)-4(n-6)=74 ?

The solutions to the equation n(4+2n)-4(n-6)=74 are n=5,n=-5

Find the zeros of n(4+2n)-4(n-6)=74

The zeros of n(4+2n)-4(n-6)=74 are n=5,n=-5

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n (4 + 2 n) - 4 (n - 6) = 74

n = 5, n = - 5

Solution steps

n (4 + 2 n) - 4 (n - 6) = 74

Expand

n (4 + 2 n) - 4 (n - 6) : 2 n^{2} + 24

2 n^{2} + 24 = 74

Move

74

to the left side

2 n^{2} - 50 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 2 ( - 50 )}{2 \cdot 2}

0^{2} - 4 \cdot 2 (- 50) = 20

n_{1, 2} = \frac{- 0 \pm 20}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- 0 + 20}{2 \cdot 2}, n_{2} = \frac{- 0 - 20}{2 \cdot 2}

n = \frac{- 0 + 20}{2 \cdot 2} : 5

n = \frac{- 0 - 20}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

n = 5, n = - 5

3 x^{2} - 22 x + 7 = 0

x, 4 x^{2} + 22 x = 12

2 x^{2} - 7 = 43

3 (x^{2} - 6) - 2 = 1

- 4 (x + 2)^{2} + 16 = 0

What are the solutions to the equation n(4+2n)-4(n-6)=74 ?
The solutions to the equation n(4+2n)-4(n-6)=74 are n=5,n=-5
Find the zeros of n(4+2n)-4(n-6)=74
The zeros of n(4+2n)-4(n-6)=74 are n=5,n=-5