What are the solutions to the equation r^2+16r+64=0 ?

The solutions to the equation r^2+16r+64=0 are r=-8

Find the zeros of r^2+16r+64=0

The zeros of r^2+16r+64=0 are r=-8

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r^{2} + 16 r + 64 = 0

r = - 8

Solution steps

r^{2} + 16 r + 64 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot 64}{2 \cdot 1}

1 6^{2} - 4 \cdot 1 \cdot 64 = 0

r_{1, 2} = \frac{- 16 \pm 0}{2 \cdot 1}

r = \frac{- 16}{2 \cdot 1}

\frac{- 16}{2 \cdot 1} = - 8

r = - 8

The solution to the quadratic equation is:

r = - 8

p^{2} - 49 = 0

(2 x + c)^{2} = 2 x - c

x, 9 x^{2} - 6 x y + y^{2} - 80 x - 40 y = 0

7 x^{2} - 2 x - 9 = 0

a^{2} = 32

What are the solutions to the equation r^2+16r+64=0 ?
The solutions to the equation r^2+16r+64=0 are r=-8
Find the zeros of r^2+16r+64=0
The zeros of r^2+16r+64=0 are r=-8