What are the solutions to the equation r^2-10r+29=0 ?

The solutions to the equation r^2-10r+29=0 are r=5+2i,r=5-2i

Find the zeros of r^2-10r+29=0

The zeros of r^2-10r+29=0 are r=5+2i,r=5-2i

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r^{2} - 10 r + 29 = 0

r = 5 + 2 i, r = 5 - 2 i

Solution steps

r^{2} - 10 r + 29 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 29}{2 \cdot 1}

Simplify

(- 10)^{2} - 4 \cdot 1 \cdot 29 : 4 i

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

Separate the solutions

r_{1} = \frac{- ( - 10 ) + 4 i}{2 \cdot 1}, r_{2} = \frac{- ( - 10 ) - 4 i}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

r = 5 + 2 i, r = 5 - 2 i

\frac{- 3}{2} + 1

(x + 3)^{2} - 6 = 19

(x - 8) (x + 2) = 0

x (8 x - 3) = 11

What are the solutions to the equation r^2-10r+29=0 ?
The solutions to the equation r^2-10r+29=0 are r=5+2i,r=5-2i
Find the zeros of r^2-10r+29=0
The zeros of r^2-10r+29=0 are r=5+2i,r=5-2i