What are the solutions to the equation 4n^2+32n+64=0 ?

The solutions to the equation 4n^2+32n+64=0 are n=-4

Find the zeros of 4n^2+32n+64=0

The zeros of 4n^2+32n+64=0 are n=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 n^{2} + 32 n + 64 = 0

n = - 4

Solution steps

4 n^{2} + 32 n + 64 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 \cdot 4 \cdot 64}{2 \cdot 4}

3 2^{2} - 4 \cdot 4 \cdot 64 = 0

n_{1, 2} = \frac{- 32 \pm 0}{2 \cdot 4}

n = \frac{- 32}{2 \cdot 4}

\frac{- 32}{2 \cdot 4} = - 4

n = - 4

The solution to the quadratic equation is:

n = - 4

5 x^{2} - 3 x - x = 0

\frac{x ^{2} - x + 5}{4} = 0

- 63 = - 2 y^{2} + 9

x, 16 x^{2} - 24 x y + 9 y^{2} - 60 x - 80 y + 100 = 0

(x + 1) (x + 2) - (x - 3) (+ 4) = 6

What are the solutions to the equation 4n^2+32n+64=0 ?
The solutions to the equation 4n^2+32n+64=0 are n=-4
Find the zeros of 4n^2+32n+64=0
The zeros of 4n^2+32n+64=0 are n=-4