What are the solutions to the equation x=x^2-12x+36 ?

The solutions to the equation x=x^2-12x+36 are x=9,x=4

Find the zeros of x=x^2-12x+36

The zeros of x=x^2-12x+36 are x=9,x=4

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x = x^{2} - 12 x + 36

x = 9, x = 4

Solution steps

x = x^{2} - 12 x + 36

Switch sides

x^{2} - 12 x + 36 = x

Move

x

to the left side

x^{2} - 13 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 9, x = 4

x^{2} - 10 x + 15 = - 7

3 x^{2} - \frac{1}{3} = 0

- 4.9 x^{2} + 2.67 x + 8.5 = 0

x^{2} - 17 x = 18

k^{2} - 5 k + 2 = 0

What are the solutions to the equation x=x^2-12x+36 ?
The solutions to the equation x=x^2-12x+36 are x=9,x=4
Find the zeros of x=x^2-12x+36
The zeros of x=x^2-12x+36 are x=9,x=4