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

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

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

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

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

x = 9, x = 4

Solution steps

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

2 x^{2} - 12 x + 14

(2 x)^{2} = 36

x^{2} + 2

x^{2} + 6 x + 114 = 0

42 x - 4.9 x^{2} = 80

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