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

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

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

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

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r^{2} + 16 r = 0

r = 0, r = - 16

Solution steps

r^{2} + 16 r = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

1 6^{2} - 4 \cdot 1 \cdot 0 = 16

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

Separate the solutions

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

r = \frac{- 16 + 16}{2 \cdot 1} : 0

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

The solutions to the quadratic equation are:

r = 0, r = - 16

∣ x + 6 ∣ < 0.001

8 x^{2} + 14 x + 3

x - 7 > - 2

x^{4} - 3 x^{3} + x^{2} - 3 x

(\frac{1000}{27})^{\frac{2}{3}}

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