What are the solutions to the equation 49x^2+126x+81=0 ?

The solutions to the equation 49x^2+126x+81=0 are x=-9/7

Find the zeros of 49x^2+126x+81=0

The zeros of 49x^2+126x+81=0 are x=-9/7

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49 x^{2} + 126 x + 81 = 0

x = - \frac{9}{7}

Decimal Notation

x = - 1.28571 \dots

Solution steps

49 x^{2} + 126 x + 81 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 126 \pm 12 6 ^{2} - 4 \cdot 49 \cdot 81}{2 \cdot 49}

12 6^{2} - 4 \cdot 49 \cdot 81 = 0

x_{1, 2} = \frac{- 126 \pm 0}{2 \cdot 49}

x = \frac{- 126}{2 \cdot 49}

\frac{- 126}{2 \cdot 49} = - \frac{9}{7}

x = - \frac{9}{7}

The solution to the quadratic equation is:

x = - \frac{9}{7}

9 x - 2 = 4 x + 9

3 x + 23 = x + 7

4 x \geq - 32

x^{2} - 2 x - 3

- 4 \cdot \frac{1}{2}

What are the solutions to the equation 49x^2+126x+81=0 ?
The solutions to the equation 49x^2+126x+81=0 are x=-9/7
Find the zeros of 49x^2+126x+81=0
The zeros of 49x^2+126x+81=0 are x=-9/7