What are the solutions to the equation 49x^2+4=28x ?

The solutions to the equation 49x^2+4=28x are x= 2/7

Find the zeros of 49x^2+4=28x

The zeros of 49x^2+4=28x are x= 2/7

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49 x^{2} + 4 = 28 x

x = \frac{2}{7}

Decimal Notation

x = 0.28571 \dots

Solution steps

49 x^{2} + 4 = 28 x

Move

28 x

to the left side

49 x^{2} + 4 - 28 x = 0

Write in the standard form

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

49 x^{2} - 28 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 49 \cdot 4}{2 \cdot 49}

(- 28)^{2} - 4 \cdot 49 \cdot 4 = 0

x_{1, 2} = \frac{- ( - 28 ) \pm 0}{2 \cdot 49}

x = \frac{- ( - 28 )}{2 \cdot 49}

\frac{- ( - 28 )}{2 \cdot 49} = \frac{2}{7}

x = \frac{2}{7}

The solution to the quadratic equation is:

x = \frac{2}{7}

x^{2} + 18 x + 81 \leq 0

4 v + 9 \leq 5

- 15 + \frac{4}{5} p = 9

2 x^{2} + 4 = 18

x^{2} - 9 x - 972 = 0

What are the solutions to the equation 49x^2+4=28x ?
The solutions to the equation 49x^2+4=28x are x= 2/7
Find the zeros of 49x^2+4=28x
The zeros of 49x^2+4=28x are x= 2/7