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

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

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

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

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

n = - \frac{9}{2}

Decimal

n = - 4.5

Solution steps

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

Solve with the quadratic formula

n_{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

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

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

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

n = - \frac{9}{2}

The solution to the quadratic equation is:

n = - \frac{9}{2}

\frac{2 \cdot 5 - 4 \cdot 5 ^{2}}{5 ^{2} - 20}

i^{69}

x^{2} + x + 12 = 0

\frac{8 + 3 i}{- 6 - 6 i}

(z + 3)^{2} = - 11

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