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

The solutions to the equation r^2+4r+36=0 are r=-2+4sqrt(2)i,r=-2-4sqrt(2)i

Find the zeros of r^2+4r+36=0

The zeros of r^2+4r+36=0 are r=-2+4sqrt(2)i,r=-2-4sqrt(2)i

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r^{2} + 4 r + 36 = 0

r = - 2 + 42 i, r = - 2 - 42 i

Solution steps

r^{2} + 4 r + 36 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 36 : 82 i

r_{1, 2} = \frac{- 4 \pm 8 2 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 4 + 8 2 i}{2 \cdot 1}, r_{2} = \frac{- 4 - 8 2 i}{2 \cdot 1}

r = \frac{- 4 + 8 2 i}{2 \cdot 1} : - 2 + 42 i

r = \frac{- 4 - 8 2 i}{2 \cdot 1} : - 2 - 42 i

The solutions to the quadratic equation are:

r = - 2 + 42 i, r = - 2 - 42 i

9 \cdot 1 0^{9}

0 = 4 x^{2} - 64 x + 192

\frac{3}{∣ 2 x - 1 ∣} > 0

\frac{3}{4} (- 1) - \frac{( - 1 ) ^{2}}{2}

- 10 b + 5 = 7 b + 5

What are the solutions to the equation r^2+4r+36=0 ?
The solutions to the equation r^2+4r+36=0 are r=-2+4sqrt(2)i,r=-2-4sqrt(2)i
Find the zeros of r^2+4r+36=0
The zeros of r^2+4r+36=0 are r=-2+4sqrt(2)i,r=-2-4sqrt(2)i