What are the solutions to the equation x^2+2x-48=(x+8) ?

The solutions to the equation x^2+2x-48=(x+8) are x=7,x=-8

Find the zeros of x^2+2x-48=(x+8)

The zeros of x^2+2x-48=(x+8) are x=7,x=-8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} + 2 x - 48 = (x + 8)

x = 7, x = - 8

Solution steps

x^{2} + 2 x - 48 = (x + 8)

Refine

x^{2} + 2 x - 48 = x + 8

Move

8

to the left side

x^{2} + 2 x - 56 = x

Move

x

to the left side

x^{2} + x - 56 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 56 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 56) = 15

x_{1, 2} = \frac{- 1 \pm 15}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 15}{2 \cdot 1}, x_{2} = \frac{- 1 - 15}{2 \cdot 1}

x = \frac{- 1 + 15}{2 \cdot 1} : 7

x = \frac{- 1 - 15}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = 7, x = - 8

0 = 30 - 4.9 x^{2}

(y - 2)^{2} = 9

8 (x - 19)^{2} = 0

3 x^{2} = 2 x^{2}

- 0.16 = 0 + \frac{1}{2} (- 9.8) t^{2}

What are the solutions to the equation x^2+2x-48=(x+8) ?
The solutions to the equation x^2+2x-48=(x+8) are x=7,x=-8
Find the zeros of x^2+2x-48=(x+8)
The zeros of x^2+2x-48=(x+8) are x=7,x=-8