What are the solutions to the equation (3n+1)^2+6(3n+1)-7=0 ?

The solutions to the equation (3n+1)^2+6(3n+1)-7=0 are n=0,n=-8/3

Find the zeros of (3n+1)^2+6(3n+1)-7=0

The zeros of (3n+1)^2+6(3n+1)-7=0 are n=0,n=-8/3

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(3 n + 1)^{2} + 6 (3 n + 1) - 7 = 0

n = 0, n = - \frac{8}{3}

Decimal Notation

n = 0, n = - 2.66666 \dots

Solution steps

(3 n + 1)^{2} + 6 (3 n + 1) - 7 = 0

Expand

(3 n + 1)^{2} + 6 (3 n + 1) - 7 : 9 n^{2} + 24 n

9 n^{2} + 24 n = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 9 \cdot 0}{2 \cdot 9}

2 4^{2} - 4 \cdot 9 \cdot 0 = 24

n_{1, 2} = \frac{- 24 \pm 24}{2 \cdot 9}

Separate the solutions

n_{1} = \frac{- 24 + 24}{2 \cdot 9}, n_{2} = \frac{- 24 - 24}{2 \cdot 9}

n = \frac{- 24 + 24}{2 \cdot 9} : 0

n = \frac{- 24 - 24}{2 \cdot 9} : - \frac{8}{3}

The solutions to the quadratic equation are:

n = 0, n = - \frac{8}{3}

- 7.4 = - 2.5 + x

(- 1)^{1}

\frac{\frac{y}{2} - \frac{1}{x}}{\frac{6 y}{x}}

- 5 x \geq 50

\frac{7 x}{2} + 5 = 8

What are the solutions to the equation (3n+1)^2+6(3n+1)-7=0 ?
The solutions to the equation (3n+1)^2+6(3n+1)-7=0 are n=0,n=-8/3
Find the zeros of (3n+1)^2+6(3n+1)-7=0
The zeros of (3n+1)^2+6(3n+1)-7=0 are n=0,n=-8/3