What is the answer to x^2+(x+1)^2=(2x-1)(x+3) ?

The answer to x^2+(x+1)^2=(2x-1)(x+3) is x= 4/3

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x^2+(x+1)^2=(2x-1)(x+3)

x^{2} + (x + 1)^{2} = (2 x - 1) (x + 3)

x = \frac{4}{3}

Decimal Notation

x = 1.33333 \dots

Solution steps

x^{2} + (x + 1)^{2} = (2 x - 1) (x + 3)

Expand

x^{2} + (x + 1)^{2} : 2 x^{2} + 2 x + 1

Expand

(2 x - 1) (x + 3) : 2 x^{2} + 5 x - 3

2 x^{2} + 2 x + 1 = 2 x^{2} + 5 x - 3

Move

1

to the right side

2 x^{2} + 2 x = 2 x^{2} + 5 x - 4

Subtract

2 x^{2} + 5 x

from both sides

2 x^{2} + 2 x - (2 x^{2} + 5 x) = 2 x^{2} + 5 x - 4 - (2 x^{2} + 5 x)

Simplify

- 3 x = - 4

Divide both sides by

- 3

x = \frac{4}{3}

3 x + 1 = x + 17

M = (x + 4)^{2} - (x - 4)^{2}

3.5 + 1.2 (6.3 - 7 x) = 9.38

(0 + 8 - 5) + k (0 + 2 + 1) = 0

\frac{2 x + 1}{5} - \frac{x + 1}{3} = 2 (\frac{3 x}{10} - (\frac{x}{6} - 1))

What is the answer to x^2+(x+1)^2=(2x-1)(x+3) ?
The answer to x^2+(x+1)^2=(2x-1)(x+3) is x= 4/3
Solve for x: x^2+(x+1)^2=(2x-1)(x+3)
The solution is x= 4/3