What are the solutions to the equation r^2+8r+16=0 ?

The solutions to the equation r^2+8r+16=0 are r=-4

Find the zeros of r^2+8r+16=0

The zeros of r^2+8r+16=0 are r=-4

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r^{2} + 8 r + 16 = 0

r = - 4

Solution steps

r^{2} + 8 r + 16 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot 16 = 0

r_{1, 2} = \frac{- 8 \pm 0}{2 \cdot 1}

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

\frac{- 8}{2 \cdot 1} = - 4

r = - 4

The solution to the quadratic equation is:

r = - 4

7 x^{2} - 175

6 x^{2} + 5 x = 6

4 x = - 4 x - 32

4 - \frac{1}{x} < 1

5 x^{2} - 8 x + 19 \geq 0

What are the solutions to the equation r^2+8r+16=0 ?
The solutions to the equation r^2+8r+16=0 are r=-4
Find the zeros of r^2+8r+16=0
The zeros of r^2+8r+16=0 are r=-4