What are the solutions to the equation 4x^2+36x+81=0 ?

The solutions to the equation 4x^2+36x+81=0 are x=-9/2

Find the zeros of 4x^2+36x+81=0

The zeros of 4x^2+36x+81=0 are x=-9/2

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4 x^{2} + 36 x + 81 = 0

x = - \frac{9}{2}

Decimal Notation

x = - 4.5

Solution steps

4 x^{2} + 36 x + 81 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 \cdot 4 \cdot 81}{2 \cdot 4}

3 6^{2} - 4 \cdot 4 \cdot 81 = 0

x_{1, 2} = \frac{- 36 \pm 0}{2 \cdot 4}

x = \frac{- 36}{2 \cdot 4}

\frac{- 36}{2 \cdot 4} = - \frac{9}{2}

x = - \frac{9}{2}

The solution to the quadratic equation is:

x = - \frac{9}{2}

2 x^{2} + 16 < 8 x

8 w^{2} - 18 w + 9

\frac{x + 14}{x - 17} \leq 1

\frac{3 t}{5} - 2 = \frac{4 t - 1}{3} + 2

\frac{1}{6} (- q + 1) = - 4 (- 3 q + 3)

What are the solutions to the equation 4x^2+36x+81=0 ?
The solutions to the equation 4x^2+36x+81=0 are x=-9/2
Find the zeros of 4x^2+36x+81=0
The zeros of 4x^2+36x+81=0 are x=-9/2