What are the solutions to the equation -x^2+4x=2x-7 ?

The solutions to the equation -x^2+4x=2x-7 are x=1-2sqrt(2),x=1+2sqrt(2)

Find the zeros of -x^2+4x=2x-7

The zeros of -x^2+4x=2x-7 are x=1-2sqrt(2),x=1+2sqrt(2)

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- x^{2} + 4 x = 2 x - 7

x = 1 - 22, x = 1 + 22

Decimal Notation

x = - 1.82842 \dots, x = 3.82842 \dots

Solution steps

- x^{2} + 4 x = 2 x - 7

Move

7

to the left side

- x^{2} + 4 x + 7 = 2 x

Move

2 x

to the left side

- x^{2} + 2 x + 7 = 0

Solve with the quadratic formula

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

2^{2} - 4 (- 1) \cdot 7 = 42

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

Separate the solutions

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

x = \frac{- 2 + 4 2}{2 ( - 1 )} : 1 - 22

x = \frac{- 2 - 4 2}{2 ( - 1 )} : 1 + 22

The solutions to the quadratic equation are:

x = 1 - 22, x = 1 + 22

x = - 3 x^{2} + 3000 x

- b^{2} = 4 b - 20

x^{2} - 6 x + 20 = - 5

x^{2} + 15 = 64

7 x + x^{2} - 126 = 0

What are the solutions to the equation -x^2+4x=2x-7 ?
The solutions to the equation -x^2+4x=2x-7 are x=1-2sqrt(2),x=1+2sqrt(2)
Find the zeros of -x^2+4x=2x-7
The zeros of -x^2+4x=2x-7 are x=1-2sqrt(2),x=1+2sqrt(2)