What are the solutions to the equation 4x-4=1-(x^2+2)/3 ?

The solutions to the equation 4x-4=1-(x^2+2)/3 are x=-13,x=1

Find the zeros of 4x-4=1-(x^2+2)/3

The zeros of 4x-4=1-(x^2+2)/3 are x=-13,x=1

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4 x - 4 = 1 - \frac{x ^{2} + 2}{3}

x = - 13, x = 1

Solution steps

4 x - 4 = 1 - \frac{x ^{2} + 2}{3}

Multiply both sides by

3

12 x - 12 = 3 - (x^{2} + 2)

Expand

3 - (x^{2} + 2) : - x^{2} + 1

12 x - 12 = - x^{2} + 1

Switch sides

- x^{2} + 1 = 12 x - 12

Move

12

to the left side

- x^{2} + 13 = 12 x

Move

12 x

to the left side

- x^{2} + 13 - 12 x = 0

Write in the standard form

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

- x^{2} - 12 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 1 ) \cdot 13}{2 ( - 1 )}

(- 12)^{2} - 4 (- 1) \cdot 13 = 14

x_{1, 2} = \frac{- ( - 12 ) \pm 14}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 14}{2 ( - 1 )}, x_{2} = \frac{- ( - 12 ) - 14}{2 ( - 1 )}

x = \frac{- ( - 12 ) + 14}{2 ( - 1 )} : - 13

x = \frac{- ( - 12 ) - 14}{2 ( - 1 )} : 1

The solutions to the quadratic equation are:

x = - 13, x = 1

5 k^{2} + 7 k + 1 = 0

- 7 n^{2} = - 488

x^{2} + 1 x - 7 = 0

x, (x - 2)^{2} = 3

3 x^{2} - x - 3

What are the solutions to the equation 4x-4=1-(x^2+2)/3 ?
The solutions to the equation 4x-4=1-(x^2+2)/3 are x=-13,x=1
Find the zeros of 4x-4=1-(x^2+2)/3
The zeros of 4x-4=1-(x^2+2)/3 are x=-13,x=1