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

The solutions to the equation 3x^2+4x=2 are x=(-2+sqrt(10))/3 ,x=-(2+sqrt(10))/3

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

The zeros of 3x^2+4x=2 are x=(-2+sqrt(10))/3 ,x=-(2+sqrt(10))/3

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3 x^{2} + 4 x = 2

x = \frac{- 2 + 10}{3}, x = - \frac{2 + 10}{3}

Decimal Notation

x = 0.38742 \dots, x = - 1.72075 \dots

Solution steps

3 x^{2} + 4 x = 2

Move

2

to the left side

3 x^{2} + 4 x - 2 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 3 (- 2) = 210

x_{1, 2} = \frac{- 4 \pm 2 10}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 4 + 2 10}{2 \cdot 3}, x_{2} = \frac{- 4 - 2 10}{2 \cdot 3}

x = \frac{- 4 + 2 10}{2 \cdot 3} : \frac{- 2 + 10}{3}

x = \frac{- 4 - 2 10}{2 \cdot 3} : - \frac{2 + 10}{3}

The solutions to the quadratic equation are:

x = \frac{- 2 + 10}{3}, x = - \frac{2 + 10}{3}

3 x^{2} - 5 x + 5 = 0

\frac{2 ( 1 - 2 i ) - 3}{( 1 - 2 i ) - 1}

- 5 x \geq 30

5 x + 12 = 0

\frac{s ^{5} t ^{3}}{s ^{7} t ^{2}}

What are the solutions to the equation 3x^2+4x=2 ?
The solutions to the equation 3x^2+4x=2 are x=(-2+sqrt(10))/3 ,x=-(2+sqrt(10))/3
Find the zeros of 3x^2+4x=2
The zeros of 3x^2+4x=2 are x=(-2+sqrt(10))/3 ,x=-(2+sqrt(10))/3