What are the solutions to the equation -x^2+6x=(x+3)2-9 ?

The solutions to the equation -x^2+6x=(x+3)2-9 are x=2-sqrt(7),x=2+sqrt(7)

Find the zeros of -x^2+6x=(x+3)2-9

The zeros of -x^2+6x=(x+3)2-9 are x=2-sqrt(7),x=2+sqrt(7)

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

x = 2 - 7, x = 2 + 7

Decimal Notation

x = - 0.64575 \dots, x = 4.64575 \dots

Solution steps

- x^{2} + 6 x = (x + 3) \cdot 2 - 9

Expand

(x + 3) \cdot 2 - 9 : 2 x - 3

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

Move

3

to the left side

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

Move

2 x

to the left side

- x^{2} + 4 x + 3 = 0

Solve with the quadratic formula

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

4^{2} - 4 (- 1) \cdot 3 = 27

x_{1, 2} = \frac{- 4 \pm 2 7}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 4 + 2 7}{2 ( - 1 )}, x_{2} = \frac{- 4 - 2 7}{2 ( - 1 )}

x = \frac{- 4 + 2 7}{2 ( - 1 )} : 2 - 7

x = \frac{- 4 - 2 7}{2 ( - 1 )} : 2 + 7

The solutions to the quadratic equation are:

x = 2 - 7, x = 2 + 7

x^{2} + 5 x - 3 = 0

3 x^{2} + 11 x + 4 = 0

4 x^{2} - 7 x - 3 = 0

t^{2} - 3 t + 3 = 0

- x^{2} - 2 x = 5

What are the solutions to the equation -x^2+6x=(x+3)2-9 ?
The solutions to the equation -x^2+6x=(x+3)2-9 are x=2-sqrt(7),x=2+sqrt(7)
Find the zeros of -x^2+6x=(x+3)2-9
The zeros of -x^2+6x=(x+3)2-9 are x=2-sqrt(7),x=2+sqrt(7)