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

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

Find the zeros of x^2=12+x

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

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x^{2} = 12 + x

x = 4, x = - 3

Solution steps

x^{2} = 12 + x

Move

x

to the left side

x^{2} - x = 12

Move

12

to the left side

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} : 4

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

The solutions to the quadratic equation are:

x = 4, x = - 3

3 (2 x - 3)^{2} - 33 = - 3

12 x^{2} - 8 x + 1 = 0

x^{2} + 8 x + 20

x, x^{2} + 2 x + 1 - m^{2} = 0

r^{2} = - 96

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