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 = 1.72075 \dots, x = - 0.38742 \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}

15 x - 20 x y

442

6 ∣ - f + 3 ∣ + 7 > 7

(x^{\frac{1}{3}})^{2}

(- 8) \frac{v}{- 8} > 2 (- 8)

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