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

The solutions to the equation 3x^2-2x=4 are x=(1+sqrt(13))/3 ,x=(1-sqrt(13))/3

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

The zeros of 3x^2-2x=4 are x=(1+sqrt(13))/3 ,x=(1-sqrt(13))/3

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3 x^{2} - 2 x = 4

x = \frac{1 + 13}{3}, x = \frac{1 - 13}{3}

Decimal Notation

x = 1.53518 \dots, x = - 0.86851 \dots

Solution steps

3 x^{2} - 2 x = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 3 (- 4) = 213

x_{1, 2} = \frac{- ( - 2 ) \pm 2 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 13}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 2 13}{2 \cdot 3}

x = \frac{- ( - 2 ) + 2 13}{2 \cdot 3} : \frac{1 + 13}{3}

x = \frac{- ( - 2 ) - 2 13}{2 \cdot 3} : \frac{1 - 13}{3}

The solutions to the quadratic equation are:

x = \frac{1 + 13}{3}, x = \frac{1 - 13}{3}

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

4 y^{2} - 5 y = 9

- 4.9 r^{2} - 120 r + 90 = 0

3 x^{2} - 8 x + 15 = 0

3 (x - 2)^{2} - 8 = 5

What are the solutions to the equation 3x^2-2x=4 ?
The solutions to the equation 3x^2-2x=4 are x=(1+sqrt(13))/3 ,x=(1-sqrt(13))/3
Find the zeros of 3x^2-2x=4
The zeros of 3x^2-2x=4 are x=(1+sqrt(13))/3 ,x=(1-sqrt(13))/3