What are the solutions to the equation r^2+18r+80=0 ?

The solutions to the equation r^2+18r+80=0 are r=-8,r=-10

Find the zeros of r^2+18r+80=0

The zeros of r^2+18r+80=0 are r=-8,r=-10

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r^{2} + 18 r + 80 = 0

r = - 8, r = - 10

Solution steps

r^{2} + 18 r + 80 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 1 \cdot 80}{2 \cdot 1}

1 8^{2} - 4 \cdot 1 \cdot 80 = 2

r_{1, 2} = \frac{- 18 \pm 2}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 18 + 2}{2 \cdot 1}, r_{2} = \frac{- 18 - 2}{2 \cdot 1}

r = \frac{- 18 + 2}{2 \cdot 1} : - 8

r = \frac{- 18 - 2}{2 \cdot 1} : - 10

The solutions to the quadratic equation are:

r = - 8, r = - 10

3 r^{2} - 16 r + 5

6 a + 5 a = - 11

x + 3 = 2 x

x (x - 3) \leq 0

y^{2} - 12 y + 11

What are the solutions to the equation r^2+18r+80=0 ?
The solutions to the equation r^2+18r+80=0 are r=-8,r=-10
Find the zeros of r^2+18r+80=0
The zeros of r^2+18r+80=0 are r=-8,r=-10