What are the solutions to the equation 81x^2+72x+16=0 ?

The solutions to the equation 81x^2+72x+16=0 are x=-4/9

Find the zeros of 81x^2+72x+16=0

The zeros of 81x^2+72x+16=0 are x=-4/9

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81 x^{2} + 72 x + 16 = 0

x = - \frac{4}{9}

Solution steps

81 x^{2} + 72 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 72 \pm 7 2 ^{2} - 4 \cdot 81 \cdot 16}{2 \cdot 81}

7 2^{2} - 4 \cdot 81 \cdot 16 = 0

x_{1, 2} = \frac{- 72 \pm 0}{2 \cdot 81}

x = \frac{- 72}{2 \cdot 81}

\frac{- 72}{2 \cdot 81} = - \frac{4}{9}

x = - \frac{4}{9}

The solution to the quadratic equation is:

x = - \frac{4}{9}

36 x^{2} - 25 = 0

\frac{( \frac{3}{2} ) ^{6}}{( \frac{3}{5} ) ^{6}} - ((\frac{5}{2})^{2})^{3}

x^{2} - 4 x - 45 \geq 0

t^{2} = 7 t

\frac{3}{5} (x + 2) = x - 4

What are the solutions to the equation 81x^2+72x+16=0 ?
The solutions to the equation 81x^2+72x+16=0 are x=-4/9
Find the zeros of 81x^2+72x+16=0
The zeros of 81x^2+72x+16=0 are x=-4/9