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}

Decimal Notation

x = - 2.25

Solution steps

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

Solve with the quadratic formula

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

7 2^{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}

x^{2} - 6 x + 9 = 0

x^{2} + 41 = 12 x + 4

e^{- l n (5)}

3 (x + 5) = 21

\frac{( 3 b ^{2} \cdot 3 b ^{3} ) ^{3}}{b ^{3}}

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