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

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

Find the zeros of x^2-12=-x

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

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x^{2} - 12 = - x

x = 3, x = - 4

Solution steps

x^{2} - 12 = - x

Move

x

to the left side

x^{2} - 12 + x = 0

Write in the standard form

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

x^{2} + x - 12 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 12) = 7

x_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 7}{2 \cdot 1}, x_{2} = \frac{- 1 - 7}{2 \cdot 1}

x = \frac{- 1 + 7}{2 \cdot 1} : 3

x = \frac{- 1 - 7}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 3, x = - 4

[(2 - x) (- 6 - x) - 5 (- 3)] = 0

3 x^{2} - 5 = 4 x

x (x + 1) - 31 = 5 (2 x + 1)

81 x^{2} + 10 = 35

0 = (x - 3)^{2} - 4

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