What are the solutions to the equation 4n^2=-6n-1 ?

The solutions to the equation 4n^2=-6n-1 are n=(-3+sqrt(5))/4 ,n=-(3+sqrt(5))/4

Find the zeros of 4n^2=-6n-1

The zeros of 4n^2=-6n-1 are n=(-3+sqrt(5))/4 ,n=-(3+sqrt(5))/4

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4 n^{2} = - 6 n - 1

n = \frac{- 3 + 5}{4}, n = - \frac{3 + 5}{4}

Solution steps

4 n^{2} = - 6 n - 1

Move

1

to the left side

4 n^{2} + 1 = - 6 n

Move

6 n

to the left side

4 n^{2} + 1 + 6 n = 0

Write in the standard form

a x^{2} + b x + c = 0

4 n^{2} + 6 n + 1 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

6^{2} - 4 \cdot 4 \cdot 1 = 25

n_{1, 2} = \frac{- 6 \pm 2 5}{2 \cdot 4}

Separate the solutions

n_{1} = \frac{- 6 + 2 5}{2 \cdot 4}, n_{2} = \frac{- 6 - 2 5}{2 \cdot 4}

n = \frac{- 6 + 2 5}{2 \cdot 4} : \frac{- 3 + 5}{4}

n = \frac{- 6 - 2 5}{2 \cdot 4} : - \frac{3 + 5}{4}

The solutions to the quadratic equation are:

n = \frac{- 3 + 5}{4}, n = - \frac{3 + 5}{4}

4 x^{2} = 16 x

80 \leq 92 + x < 90

\frac{x}{2} = 5 + \frac{x - 40}{5}

8 x + 6 y

- (\frac{4}{3})^{2}

What are the solutions to the equation 4n^2=-6n-1 ?
The solutions to the equation 4n^2=-6n-1 are n=(-3+sqrt(5))/4 ,n=-(3+sqrt(5))/4
Find the zeros of 4n^2=-6n-1
The zeros of 4n^2=-6n-1 are n=(-3+sqrt(5))/4 ,n=-(3+sqrt(5))/4