What are the solutions to the equation -4x^2=48x+128 ?

The solutions to the equation -4x^2=48x+128 are x=-8,x=-4

Find the zeros of -4x^2=48x+128

The zeros of -4x^2=48x+128 are x=-8,x=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 4 x^{2} = 48 x + 128

x = - 8, x = - 4

Solution steps

- 4 x^{2} = 48 x + 128

Move

128

to the left side

- 4 x^{2} - 128 = 48 x

Move

48 x

to the left side

- 4 x^{2} - 128 - 48 x = 0

Write in the standard form

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

- 4 x^{2} - 48 x - 128 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 ( - 4 ) ( - 128 )}{2 ( - 4 )}

(- 48)^{2} - 4 (- 4) (- 128) = 16

x_{1, 2} = \frac{- ( - 48 ) \pm 16}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 48 ) + 16}{2 ( - 4 )}, x_{2} = \frac{- ( - 48 ) - 16}{2 ( - 4 )}

x = \frac{- ( - 48 ) + 16}{2 ( - 4 )} : - 8

x = \frac{- ( - 48 ) - 16}{2 ( - 4 )} : - 4

The solutions to the quadratic equation are:

x = - 8, x = - 4

4 w^{2} = 4 w + 3

x^{2} + 12 x - 69

3 x^{2} = 12 x^{2} - 144

3 (y + 1)^{2} - 10 (y + 1) + 8 = 0

3 w^{2} + 49 w + 60 = 0

What are the solutions to the equation -4x^2=48x+128 ?
The solutions to the equation -4x^2=48x+128 are x=-8,x=-4
Find the zeros of -4x^2=48x+128
The zeros of -4x^2=48x+128 are x=-8,x=-4