What are the solutions to the equation -x^2-2x+48=3x+12 ?

The solutions to the equation -x^2-2x+48=3x+12 are x=-9,x=4

Find the zeros of -x^2-2x+48=3x+12

The zeros of -x^2-2x+48=3x+12 are x=-9,x=4

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 = 3 x + 12

x = - 9, x = 4

Solution steps

- x^{2} - 2 x + 48 = 3 x + 12

Move

12

to the left side

- x^{2} - 2 x + 36 = 3 x

Move

3 x

to the left side

- x^{2} - 5 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 1 ) \cdot 36}{2 ( - 1 )}

(- 5)^{2} - 4 (- 1) \cdot 36 = 13

x_{1, 2} = \frac{- ( - 5 ) \pm 13}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 13}{2 ( - 1 )}, x_{2} = \frac{- ( - 5 ) - 13}{2 ( - 1 )}

x = \frac{- ( - 5 ) + 13}{2 ( - 1 )} : - 9

x = \frac{- ( - 5 ) - 13}{2 ( - 1 )} : 4

The solutions to the quadratic equation are:

x = - 9, x = 4

9 x^{2} - 3 = 0

9 (x^{2} - 11) - 8 = 14

4 x^{2} - 8 = x

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

x^{2} - 1 = 24

What are the solutions to the equation -x^2-2x+48=3x+12 ?
The solutions to the equation -x^2-2x+48=3x+12 are x=-9,x=4
Find the zeros of -x^2-2x+48=3x+12
The zeros of -x^2-2x+48=3x+12 are x=-9,x=4