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

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

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

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

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3 x^{2} - 4 x = 3

x = \frac{2 + 13}{3}, x = \frac{2 - 13}{3}

Solution steps

3 x^{2} - 4 x = 3

Move

3

to the left side

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

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- ( - 4 ) \pm 2 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 13}{2 \cdot 3}, x_{2} = \frac{- ( - 4 ) - 2 13}{2 \cdot 3}

x = \frac{- ( - 4 ) + 2 13}{2 \cdot 3} : \frac{2 + 13}{3}

x = \frac{- ( - 4 ) - 2 13}{2 \cdot 3} : \frac{2 - 13}{3}

The solutions to the quadratic equation are:

x = \frac{2 + 13}{3}, x = \frac{2 - 13}{3}

x^{2} + 4 = 6 x - 5

49 x^{2} - 21 x + 2 = 0

2 x^{2} = \frac{a ^{3}}{2}

t (6 - t) = 0

175 - 50 x + x^{2} = 0

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