What are the solutions to the equation (2x+1)3x+6x(2x+1)=-9x(1+2x) ?

The solutions to the equation (2x+1)3x+6x(2x+1)=-9x(1+2x) are x=0,x=-1/2

Find the zeros of (2x+1)3x+6x(2x+1)=-9x(1+2x)

The zeros of (2x+1)3x+6x(2x+1)=-9x(1+2x) are x=0,x=-1/2

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(2 x + 1) 3 x + 6 x (2 x + 1) = - 9 x (1 + 2 x)

x = 0, x = - \frac{1}{2}

Decimal Notation

x = 0, x = - 0.5

Solution steps

(2 x + 1) \cdot 3 x + 6 x (2 x + 1) = - 9 x (1 + 2 x)

Move

9 x (1 + 2 x)

to the left side

18 x (2 x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 2 x + 1 = 0

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

The solutions to the quadratic equation are:

x = 0, x = - \frac{1}{2}

x, (C - x) (1 + k x) = B (k q x)

10 x^{2} + 7 x - 5 = 0

0 = - 2 x^{2} - 16 x - 30

4 x^{2} + 30 x - 54 = 0

- 12 x + x^{2} = 28

What are the solutions to the equation (2x+1)3x+6x(2x+1)=-9x(1+2x) ?
The solutions to the equation (2x+1)3x+6x(2x+1)=-9x(1+2x) are x=0,x=-1/2
Find the zeros of (2x+1)3x+6x(2x+1)=-9x(1+2x)
The zeros of (2x+1)3x+6x(2x+1)=-9x(1+2x) are x=0,x=-1/2