What are the solutions to the equation -x^2+4x=-4x+3 ?

The solutions to the equation -x^2+4x=-4x+3 are x=4-sqrt(13),x=4+sqrt(13)

Find the zeros of -x^2+4x=-4x+3

The zeros of -x^2+4x=-4x+3 are x=4-sqrt(13),x=4+sqrt(13)

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

x = 4 - 13, x = 4 + 13

Decimal Notation

x = 0.39444 \dots, x = 7.60555 \dots

Solution steps

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

Move

3

to the left side

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

Move

4 x

to the left side

- x^{2} + 8 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 1 ) ( - 3 )}{2 ( - 1 )}

8^{2} - 4 (- 1) (- 3) = 213

x_{1, 2} = \frac{- 8 \pm 2 13}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 8 + 2 13}{2 ( - 1 )}, x_{2} = \frac{- 8 - 2 13}{2 ( - 1 )}

x = \frac{- 8 + 2 13}{2 ( - 1 )} : 4 - 13

x = \frac{- 8 - 2 13}{2 ( - 1 )} : 4 + 13

The solutions to the quadratic equation are:

x = 4 - 13, x = 4 + 13

- 9 x^{2} + 3 x + 10 = 0

9 p^{2} - 7 = 9 p

- 7 x^{2} + 12 x + 64 = 0

- 3 x^{2} - 9 = - 5 x

10 x^{2} - 50 = 0

What are the solutions to the equation -x^2+4x=-4x+3 ?
The solutions to the equation -x^2+4x=-4x+3 are x=4-sqrt(13),x=4+sqrt(13)
Find the zeros of -x^2+4x=-4x+3
The zeros of -x^2+4x=-4x+3 are x=4-sqrt(13),x=4+sqrt(13)