What are the solutions to the equation r^2+200r+10=0 ?

The solutions to the equation r^2+200r+10=0 are r=-100+3sqrt(1110),r=-100-3sqrt(1110)

Find the zeros of r^2+200r+10=0

The zeros of r^2+200r+10=0 are r=-100+3sqrt(1110),r=-100-3sqrt(1110)

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r^{2} + 200 r + 10 = 0

r = - 100 + 31110, r = - 100 - 31110

Decimal Notation

r = - 0.05001 \dots, r = - 199.94998 \dots

Solution steps

r^{2} + 200 r + 10 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 200 \pm 20 0 ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

20 0^{2} - 4 \cdot 1 \cdot 10 = 61110

r_{1, 2} = \frac{- 200 \pm 6 1110}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 200 + 6 1110}{2 \cdot 1}, r_{2} = \frac{- 200 - 6 1110}{2 \cdot 1}

r = \frac{- 200 + 6 1110}{2 \cdot 1} : - 100 + 31110

r = \frac{- 200 - 6 1110}{2 \cdot 1} : - 100 - 31110

The solutions to the quadratic equation are:

r = - 100 + 31110, r = - 100 - 31110

(a - b) \cdot z + 3 \cdot (b - a)

4 h^{2} + 4 h - 3

9 - ∣ 3 x + 9 ∣ = 6

p^{2} r^{2} - 100 x^{2}

25 x^{2} (2 x^{2} + 3) - (2 x^{2} + 3)

What are the solutions to the equation r^2+200r+10=0 ?
The solutions to the equation r^2+200r+10=0 are r=-100+3sqrt(1110),r=-100-3sqrt(1110)
Find the zeros of r^2+200r+10=0
The zeros of r^2+200r+10=0 are r=-100+3sqrt(1110),r=-100-3sqrt(1110)