What are the solutions to the equation (2n+1)^2+3(2n+1)-4=0 ?

The solutions to the equation (2n+1)^2+3(2n+1)-4=0 are n=0,n=-5/2

Find the zeros of (2n+1)^2+3(2n+1)-4=0

The zeros of (2n+1)^2+3(2n+1)-4=0 are n=0,n=-5/2

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(2 n + 1)^{2} + 3 (2 n + 1) - 4 = 0

n = 0, n = - \frac{5}{2}

Decimal Notation

n = 0, n = - 2.5

Solution steps

(2 n + 1)^{2} + 3 (2 n + 1) - 4 = 0

Expand

(2 n + 1)^{2} + 3 (2 n + 1) - 4 : 4 n^{2} + 10 n

4 n^{2} + 10 n = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

1 0^{2} - 4 \cdot 4 \cdot 0 = 10

n_{1, 2} = \frac{- 10 \pm 10}{2 \cdot 4}

Separate the solutions

n_{1} = \frac{- 10 + 10}{2 \cdot 4}, n_{2} = \frac{- 10 - 10}{2 \cdot 4}

n = \frac{- 10 + 10}{2 \cdot 4} : 0

n = \frac{- 10 - 10}{2 \cdot 4} : - \frac{5}{2}

The solutions to the quadratic equation are:

n = 0, n = - \frac{5}{2}

8 x^{2} + 20 x + 8

- 9 r^{2} + 1

x^{3} - 5 x^{2} + 4 x + 10

6 x - 9 = 7 - 2 x

4 x^{2} y - 8 x y^{3}

What are the solutions to the equation (2n+1)^2+3(2n+1)-4=0 ?
The solutions to the equation (2n+1)^2+3(2n+1)-4=0 are n=0,n=-5/2
Find the zeros of (2n+1)^2+3(2n+1)-4=0
The zeros of (2n+1)^2+3(2n+1)-4=0 are n=0,n=-5/2