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

The solutions to the equation -2(x-3)^2=2x-9 are x=(5-sqrt(7))/2 ,x=(5+sqrt(7))/2

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

The zeros of -2(x-3)^2=2x-9 are x=(5-sqrt(7))/2 ,x=(5+sqrt(7))/2

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

x = \frac{5 - 7}{2}, x = \frac{5 + 7}{2}

Decimal Notation

x = 1.17712 \dots, x = 3.82287 \dots

Solution steps

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

Expand

- 2 (x - 3)^{2} : - 2 x^{2} + 12 x - 18

- 2 x^{2} + 12 x - 18 = 2 x - 9

Move

9

to the left side

- 2 x^{2} + 12 x - 9 = 2 x

Move

2 x

to the left side

- 2 x^{2} + 10 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 2 ) ( - 9 )}{2 ( - 2 )}

1 0^{2} - 4 (- 2) (- 9) = 27

x_{1, 2} = \frac{- 10 \pm 2 7}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 10 + 2 7}{2 ( - 2 )}, x_{2} = \frac{- 10 - 2 7}{2 ( - 2 )}

x = \frac{- 10 + 2 7}{2 ( - 2 )} : \frac{5 - 7}{2}

x = \frac{- 10 - 2 7}{2 ( - 2 )} : \frac{5 + 7}{2}

The solutions to the quadratic equation are:

x = \frac{5 - 7}{2}, x = \frac{5 + 7}{2}

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

- 71 + 4 x \leq - 87

4! \cdot 4!

x + \frac{13}{12} = \frac{11}{12} x

4 > 7 x + 9

What are the solutions to the equation -2(x-3)^2=2x-9 ?
The solutions to the equation -2(x-3)^2=2x-9 are x=(5-sqrt(7))/2 ,x=(5+sqrt(7))/2
Find the zeros of -2(x-3)^2=2x-9
The zeros of -2(x-3)^2=2x-9 are x=(5-sqrt(7))/2 ,x=(5+sqrt(7))/2