What are the solutions to the equation r^2-8r+52=0 ?

The solutions to the equation r^2-8r+52=0 are r=4+6i,r=4-6i

Find the zeros of r^2-8r+52=0

The zeros of r^2-8r+52=0 are r=4+6i,r=4-6i

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r^{2} - 8 r + 52 = 0

r = 4 + 6 i, r = 4 - 6 i

Solution steps

r^{2} - 8 r + 52 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 52}{2 \cdot 1}

Simplify

(- 8)^{2} - 4 \cdot 1 \cdot 52 : 12 i

r_{1, 2} = \frac{- ( - 8 ) \pm 12 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 8 ) + 12 i}{2 \cdot 1}, r_{2} = \frac{- ( - 8 ) - 12 i}{2 \cdot 1}

r = \frac{- ( - 8 ) + 12 i}{2 \cdot 1} : 4 + 6 i

r = \frac{- ( - 8 ) - 12 i}{2 \cdot 1} : 4 - 6 i

The solutions to the quadratic equation are:

r = 4 + 6 i, r = 4 - 6 i

6 \cdot (b + 1) = - 6 b

\frac{16 !}{( 16 - 7 ) ! 7 !} 0. 6^{7} 0. 4^{9}

x^{2} + 1 \leq 0

32 - 32 x = (6 x + 4) (x - 1)

\frac{86204}{461.4 \cdot 294.2}

What are the solutions to the equation r^2-8r+52=0 ?
The solutions to the equation r^2-8r+52=0 are r=4+6i,r=4-6i
Find the zeros of r^2-8r+52=0
The zeros of r^2-8r+52=0 are r=4+6i,r=4-6i