What are the solutions to the equation r^2+10r+26=0 ?

The solutions to the equation r^2+10r+26=0 are r=-5+i,r=-5-i

Find the zeros of r^2+10r+26=0

The zeros of r^2+10r+26=0 are r=-5+i,r=-5-i

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r^{2} + 10 r + 26 = 0

r = - 5 + i, r = - 5 - i

Solution steps

r^{2} + 10 r + 26 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot 26}{2 \cdot 1}

Simplify

1 0^{2} - 4 \cdot 1 \cdot 26 : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

r = - 5 + i, r = - 5 - i

x^{2} - 10 x y - 24 y^{2}

- 9 - - 4

- \frac{9}{2} + 8

f = 3^{2} - 3

x + 1 > ∣ x - 3 ∣

What are the solutions to the equation r^2+10r+26=0 ?
The solutions to the equation r^2+10r+26=0 are r=-5+i,r=-5-i
Find the zeros of r^2+10r+26=0
The zeros of r^2+10r+26=0 are r=-5+i,r=-5-i