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

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

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

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

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3 x^{2} - 4 = 2 x

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} - 4 = 2 x

Move

2 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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}

y^{2} = 225

4 x (x - 3) = 6 - 2 x

C, X^{2} + Y^{2} = C^{2}

y^{2} = 11 y

y^{2} = 125

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