What are the solutions to the equation 672= n/2 (4+(n-1)3) ?

The solutions to the equation 672= n/2 (4+(n-1)3) are n=21,n=-64/3

Find the zeros of 672= n/2 (4+(n-1)3)

The zeros of 672= n/2 (4+(n-1)3) are n=21,n=-64/3

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672 = \frac{n}{2} (4 + (n - 1) 3)

n = 21, n = - \frac{64}{3}

Decimal Notation

n = 21, n = - 21.33333 \dots

Solution steps

672 = \frac{n}{2} (4 + (n - 1) \cdot 3)

Multiply both sides by

2

1344 = n (4 + 3 (n - 1))

Expand

n (4 + 3 (n - 1)) : 3 n^{2} + n

1344 = 3 n^{2} + n

Switch sides

3 n^{2} + n = 1344

Move

1344

to the left side

3 n^{2} + n - 1344 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 1344 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 1344) = 127

n_{1, 2} = \frac{- 1 \pm 127}{2 \cdot 3}

Separate the solutions

n_{1} = \frac{- 1 + 127}{2 \cdot 3}, n_{2} = \frac{- 1 - 127}{2 \cdot 3}

n = \frac{- 1 + 127}{2 \cdot 3} : 21

n = \frac{- 1 - 127}{2 \cdot 3} : - \frac{64}{3}

The solutions to the quadratic equation are:

n = 21, n = - \frac{64}{3}

x, x^{2} + 6 x - 10 = 0

- (x + 7) - 2 x (x - 3) = - 4 x

2 x^{2} - 14 x = 21

7 x^{2} + 8 x + 2 = 0

3 x^{2} - 12 = (x - 4)^{2} + 8 x

What are the solutions to the equation 672= n/2 (4+(n-1)3) ?
The solutions to the equation 672= n/2 (4+(n-1)3) are n=21,n=-64/3
Find the zeros of 672= n/2 (4+(n-1)3)
The zeros of 672= n/2 (4+(n-1)3) are n=21,n=-64/3