What are the solutions to the equation (2x)/7 =2x^2 ?

The solutions to the equation (2x)/7 =2x^2 are x= 1/7 ,x=0

Find the zeros of (2x)/7 =2x^2

The zeros of (2x)/7 =2x^2 are x= 1/7 ,x=0

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\frac{2 x}{7} = 2 x^{2}

x = \frac{1}{7}, x = 0

Solution steps

\frac{2 x}{7} = 2 x^{2}

Multiply both sides by

7

2 x = 14 x^{2}

Switch sides

14 x^{2} = 2 x

Move

2 x

to the left side

14 x^{2} - 2 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 14 \cdot 0}{2 \cdot 14}

(- 2)^{2} - 4 \cdot 14 \cdot 0 = 2

x_{1, 2} = \frac{- ( - 2 ) \pm 2}{2 \cdot 14}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2}{2 \cdot 14}, x_{2} = \frac{- ( - 2 ) - 2}{2 \cdot 14}

x = \frac{- ( - 2 ) + 2}{2 \cdot 14} : \frac{1}{7}

x = \frac{- ( - 2 ) - 2}{2 \cdot 14} : 0

The solutions to the quadratic equation are:

x = \frac{1}{7}, x = 0

4 v^{2} + 720 = 0

2 x - 4 = (x + 1) (2 x - 4) - 5 (x + 1)

10 y^{2} - 25 y = 0

x^{2} = 7^{2} + 8^{2}

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

What are the solutions to the equation (2x)/7 =2x^2 ?
The solutions to the equation (2x)/7 =2x^2 are x= 1/7 ,x=0
Find the zeros of (2x)/7 =2x^2
The zeros of (2x)/7 =2x^2 are x= 1/7 ,x=0