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

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

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

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

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

x = \frac{9}{4}

Solution steps

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

Solve with the quadratic formula

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

(- 72)^{2} - 4 \cdot 16 \cdot 81 = 0

x_{1, 2} = \frac{- ( - 72 ) \pm 0}{2 \cdot 16}

x = \frac{- ( - 72 )}{2 \cdot 16}

\frac{- ( - 72 )}{2 \cdot 16} = \frac{9}{4}

x = \frac{9}{4}

The solution to the quadratic equation is:

x = \frac{9}{4}

(\frac{1}{7})^{- 3}

4^{x} < 8

2 (12) + 24 = 180

(\frac{2 + - 3}{2}) (\frac{2 - - 3}{2})

\frac{\frac{1}{x}}{\frac{2}{y}}

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