What are the solutions to the equation 1=3x^2+2x ?

The solutions to the equation 1=3x^2+2x are x= 1/3 ,x=-1

Find the zeros of 1=3x^2+2x

The zeros of 1=3x^2+2x are x= 1/3 ,x=-1

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1 = 3 x^{2} + 2 x

x = \frac{1}{3}, x = - 1

Solution steps

1 = 3 x^{2} + 2 x

Switch sides

3 x^{2} + 2 x = 1

Move

1

to the left side

3 x^{2} + 2 x - 1 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 3 (- 1) = 4

x_{1, 2} = \frac{- 2 \pm 4}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 2 + 4}{2 \cdot 3}, x_{2} = \frac{- 2 - 4}{2 \cdot 3}

x = \frac{- 2 + 4}{2 \cdot 3} : \frac{1}{3}

x = \frac{- 2 - 4}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

x = \frac{1}{3}, x = - 1

\frac{2}{3} \cdot \frac{9}{4}

4 (3 d - 2) = 8 d - 5

16 m^{2} - p^{2}

20 a^{2} - 45

3 x^{2} + 21 = 0

What are the solutions to the equation 1=3x^2+2x ?
The solutions to the equation 1=3x^2+2x are x= 1/3 ,x=-1
Find the zeros of 1=3x^2+2x
The zeros of 1=3x^2+2x are x= 1/3 ,x=-1