What are the solutions to the equation 81r^2+180r+100=0 ?

The solutions to the equation 81r^2+180r+100=0 are r=-10/9

Find the zeros of 81r^2+180r+100=0

The zeros of 81r^2+180r+100=0 are r=-10/9

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81 r^{2} + 180 r + 100 = 0

r = - \frac{10}{9}

Solution steps

81 r^{2} + 180 r + 100 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 180 \pm 18 0 ^{2} - 4 \cdot 81 \cdot 100}{2 \cdot 81}

18 0^{2} - 4 \cdot 81 \cdot 100 = 0

r_{1, 2} = \frac{- 180 \pm 0}{2 \cdot 81}

r = \frac{- 180}{2 \cdot 81}

\frac{- 180}{2 \cdot 81} = - \frac{10}{9}

r = - \frac{10}{9}

The solution to the quadratic equation is:

r = - \frac{10}{9}

6 a^{2} - 13 a + 6

- 2 + \frac{1}{4}

7 x^{2} + 6 x - 1

2 \cdot 1 0^{5}

12 p^{3} - 21 p^{2} + 28 p - 49

What are the solutions to the equation 81r^2+180r+100=0 ?
The solutions to the equation 81r^2+180r+100=0 are r=-10/9
Find the zeros of 81r^2+180r+100=0
The zeros of 81r^2+180r+100=0 are r=-10/9