What are the solutions to the equation 49x^2+28x+4=0 ?

The solutions to the equation 49x^2+28x+4=0 are x=-2/7

Find the zeros of 49x^2+28x+4=0

The zeros of 49x^2+28x+4=0 are x=-2/7

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

x = - \frac{2}{7}

Decimal Notation

x = - 0.28571 \dots

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 49 \cdot 4}{2 \cdot 49}

2 8^{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}

3^{\frac{1}{4}} \cdot 2 7^{\frac{1}{4}}

x^{2} - 5 x - 12 = 0

x, f (a + h) = 3 x^{2} + 2 x

\frac{x ^{2} + 9 x + 20}{x + 1} > 0

\frac{x ^{5} y ^{4}}{x ^{- 5} y ^{4}}

What are the solutions to the equation 49x^2+28x+4=0 ?
The solutions to the equation 49x^2+28x+4=0 are x=-2/7
Find the zeros of 49x^2+28x+4=0
The zeros of 49x^2+28x+4=0 are x=-2/7