What are the solutions to the equation -4x^2+32x+80=0 ?

The solutions to the equation -4x^2+32x+80=0 are x=-2,x=10

Find the zeros of -4x^2+32x+80=0

The zeros of -4x^2+32x+80=0 are x=-2,x=10

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 4 x^{2} + 32 x + 80 = 0

x = - 2, x = 10

Solution steps

- 4 x^{2} + 32 x + 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 4 ) \cdot 80}{2 ( - 4 )}

3 2^{2} - 4 (- 4) \cdot 80 = 48

x_{1, 2} = \frac{- 32 \pm 48}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 32 + 48}{2 ( - 4 )}, x_{2} = \frac{- 32 - 48}{2 ( - 4 )}

x = \frac{- 32 + 48}{2 ( - 4 )} : - 2

x = \frac{- 32 - 48}{2 ( - 4 )} : 10

The solutions to the quadratic equation are:

x = - 2, x = 10

x^{2} + 8 x = - 12 + 0

- 2 x^{2} + 5 x + 8 = 0

5 x^{2} + 29 x + 20 = 0

y = x^{2} - 9 x + 20, x

(x + 1) (8 x + 1) = 18 x

What are the solutions to the equation -4x^2+32x+80=0 ?
The solutions to the equation -4x^2+32x+80=0 are x=-2,x=10
Find the zeros of -4x^2+32x+80=0
The zeros of -4x^2+32x+80=0 are x=-2,x=10