What are the solutions to the equation 3x=10x^2-2 ?

The solutions to the equation 3x=10x^2-2 are x=(3+sqrt(89))/(20),x=(3-sqrt(89))/(20)

Find the zeros of 3x=10x^2-2

The zeros of 3x=10x^2-2 are x=(3+sqrt(89))/(20),x=(3-sqrt(89))/(20)

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3 x = 10 x^{2} - 2

x = \frac{3 + 89}{20}, x = \frac{3 - 89}{20}

Decimal Notation

x = 0.62169 \dots, x = - 0.32169 \dots

Solution steps

3 x = 10 x^{2} - 2

Switch sides

10 x^{2} - 2 = 3 x

Move

3 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 10 (- 2) = 89

x_{1, 2} = \frac{- ( - 3 ) \pm 89}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 89}{2 \cdot 10}, x_{2} = \frac{- ( - 3 ) - 89}{2 \cdot 10}

x = \frac{- ( - 3 ) + 89}{2 \cdot 10} : \frac{3 + 89}{20}

x = \frac{- ( - 3 ) - 89}{2 \cdot 10} : \frac{3 - 89}{20}

The solutions to the quadratic equation are:

x = \frac{3 + 89}{20}, x = \frac{3 - 89}{20}

b^{2} + b = 2

x^{2} + 11 x

y^{2} - 4 y - 5 = 0

x^{2} + 16 x + 24 = 0

∣ x ∣ = 2

What are the solutions to the equation 3x=10x^2-2 ?
The solutions to the equation 3x=10x^2-2 are x=(3+sqrt(89))/(20),x=(3-sqrt(89))/(20)
Find the zeros of 3x=10x^2-2
The zeros of 3x=10x^2-2 are x=(3+sqrt(89))/(20),x=(3-sqrt(89))/(20)