What is the answer to 3x+sqrt(2x^2+4x)=2 ?

The answer to 3x+sqrt(2x^2+4x)=2 is x= 2/7

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3x+sqrt(2x^2+4x)=2

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

x = \frac{2}{7}

Decimal Notation

x = 0.28571 \dots

Solution steps

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

Remove square roots

2 x^{2} + 4 x = 4 - 12 x + 9 x^{2}

Solve

2 x^{2} + 4 x = 4 - 12 x + 9 x^{2} : x = 2, x = \frac{2}{7}

x = 2, x = \frac{2}{7}

Verify Solutions:

x = 2

False

, x = \frac{2}{7}

True

The solution is

x = \frac{2}{7}

2 x + 3 + x - 7 = 3

(x) = x - 1

x - 1 - 3 = x + 2

y, x = 4 - y

What is the answer to 3x+sqrt(2x^2+4x)=2 ?
The answer to 3x+sqrt(2x^2+4x)=2 is x= 2/7
Solve for x: 3x+sqrt(2x^2+4x)=2
The solution is x= 2/7