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

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

Find the zeros of 12=x^2-x

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

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12 = x^{2} - x

x = 4, x = - 3

Solution steps

12 = x^{2} - x

Switch sides

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

a, p (a - 4) = (a - 4)^{2} - 5

x, x^{2} + 8 x + 1 = 0

6 x = 2 x^{2} + 3

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

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

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