What are the solutions to the equation (n-9)(4n+1)=0 ?

The solutions to the equation (n-9)(4n+1)=0 are n=9,n=-1/4

Find the zeros of (n-9)(4n+1)=0

The zeros of (n-9)(4n+1)=0 are n=9,n=-1/4

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(n - 9) (4 n + 1) = 0

n = 9, n = - \frac{1}{4}

Decimal Notation

n = 9, n = - 0.25

Solution steps

(n - 9) (4 n + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

n - 9 = 0 or 4 n + 1 = 0

Solve

n - 9 = 0 : n = 9

Solve

4 n + 1 = 0 : n = - \frac{1}{4}

The solutions to the quadratic equation are:

n = 9, n = - \frac{1}{4}

3 x^{2} - 10 x - 9 = 0

3 x y^{2} - 27 x^{3}

64 x^{2} y^{2} - 144 z^{2}

- 8 \leq z + 6.4

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

What are the solutions to the equation (n-9)(4n+1)=0 ?
The solutions to the equation (n-9)(4n+1)=0 are n=9,n=-1/4
Find the zeros of (n-9)(4n+1)=0
The zeros of (n-9)(4n+1)=0 are n=9,n=-1/4