What are the solutions to the equation r^2+10r+22=0 ?

The solutions to the equation r^2+10r+22=0 are r=-5+sqrt(3),r=-5-sqrt(3)

Find the zeros of r^2+10r+22=0

The zeros of r^2+10r+22=0 are r=-5+sqrt(3),r=-5-sqrt(3)

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r^{2} + 10 r + 22 = 0

r = - 5 + 3, r = - 5 - 3

Decimal Notation

r = - 3.26794 \dots, r = - 6.73205 \dots

Solution steps

r^{2} + 10 r + 22 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot 22}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot 22 = 23

r_{1, 2} = \frac{- 10 \pm 2 3}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 10 + 2 3}{2 \cdot 1}, r_{2} = \frac{- 10 - 2 3}{2 \cdot 1}

r = \frac{- 10 + 2 3}{2 \cdot 1} : - 5 + 3

r = \frac{- 10 - 2 3}{2 \cdot 1} : - 5 - 3

The solutions to the quadratic equation are:

r = - 5 + 3, r = - 5 - 3

2 x^{2} + 15 x - 11 = - 3

21 c^{2} - 16 c - 6 = 1 - 2 c

- \frac{40}{3} = - \frac{10}{3} x^{2}

(x - 11)^{2} - 81 = 0

4800008 x^{2} = 0.0233 - x

What are the solutions to the equation r^2+10r+22=0 ?
The solutions to the equation r^2+10r+22=0 are r=-5+sqrt(3),r=-5-sqrt(3)
Find the zeros of r^2+10r+22=0
The zeros of r^2+10r+22=0 are r=-5+sqrt(3),r=-5-sqrt(3)