What are the solutions to the equation x^2+9=2x(3-x)+9 ?

The solutions to the equation x^2+9=2x(3-x)+9 are x=0,x=2

Find the zeros of x^2+9=2x(3-x)+9

The zeros of x^2+9=2x(3-x)+9 are x=0,x=2

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x^{2} + 9 = 2 x (3 - x) + 9

x = 0, x = 2

Solution steps

x^{2} + 9 = 2 x (3 - x) + 9

Expand

2 x (3 - x) + 9 : 6 x - 2 x^{2} + 9

x^{2} + 9 = 6 x - 2 x^{2} + 9

Switch sides

6 x - 2 x^{2} + 9 = x^{2} + 9

Move

9

to the left side

- 2 x^{2} + 6 x = x^{2}

Move

x^{2}

to the left side

- 3 x^{2} + 6 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

6^{2} - 4 (- 3) \cdot 0 = 6

x_{1, 2} = \frac{- 6 \pm 6}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 6 + 6}{2 ( - 3 )}, x_{2} = \frac{- 6 - 6}{2 ( - 3 )}

x = \frac{- 6 + 6}{2 ( - 3 )} : 0

x = \frac{- 6 - 6}{2 ( - 3 )} : 2

The solutions to the quadratic equation are:

x = 0, x = 2

x, f (x + 1) = 2 x^{2}

r^{2} - 10 r = - 17

- 9 x^{2} + 1 = 0

x^{2} = 1225

x, P (x - Q)^{2} + R = 0

What are the solutions to the equation x^2+9=2x(3-x)+9 ?
The solutions to the equation x^2+9=2x(3-x)+9 are x=0,x=2
Find the zeros of x^2+9=2x(3-x)+9
The zeros of x^2+9=2x(3-x)+9 are x=0,x=2