What are the solutions to the equation r^2-10r+41=0 ?

The solutions to the equation r^2-10r+41=0 are r=5+4i,r=5-4i

Find the zeros of r^2-10r+41=0

The zeros of r^2-10r+41=0 are r=5+4i,r=5-4i

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r^{2} - 10 r + 41 = 0

r = 5 + 4 i, r = 5 - 4 i

Solution steps

r^{2} - 10 r + 41 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 41}{2 \cdot 1}

Simplify

(- 10)^{2} - 4 \cdot 1 \cdot 41 : 8 i

r_{1, 2} = \frac{- ( - 10 ) \pm 8 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 10 ) + 8 i}{2 \cdot 1}, r_{2} = \frac{- ( - 10 ) - 8 i}{2 \cdot 1}

r = \frac{- ( - 10 ) + 8 i}{2 \cdot 1} : 5 + 4 i

r = \frac{- ( - 10 ) - 8 i}{2 \cdot 1} : 5 - 4 i

The solutions to the quadratic equation are:

r = 5 + 4 i, r = 5 - 4 i

- 5 v^{2} - 8 v + 4

- x^{2} + 4 x = x^{2} - 2 x

\frac{1}{3} x - 3 = - 2 x + 6

3 x^{- 2}

8 - p = - 12 - 3 p

What are the solutions to the equation r^2-10r+41=0 ?
The solutions to the equation r^2-10r+41=0 are r=5+4i,r=5-4i
Find the zeros of r^2-10r+41=0
The zeros of r^2-10r+41=0 are r=5+4i,r=5-4i