What are the solutions to the equation r^2-12r+52=0 ?

The solutions to the equation r^2-12r+52=0 are r=6+4i,r=6-4i

Find the zeros of r^2-12r+52=0

The zeros of r^2-12r+52=0 are r=6+4i,r=6-4i

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r^{2} - 12 r + 52 = 0

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

Solution steps

r^{2} - 12 r + 52 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 52}{2 \cdot 1}

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

- x^{2} + 5 x

x^{2} - 87 x + 1180 = 0

t^{2} + 22 t = - 27

(3 n - 1) \cdot x^{2} - (5 n + 2) \cdot x + 2 = 0

356 = 500 - 4 q^{2}

What are the solutions to the equation r^2-12r+52=0 ?
The solutions to the equation r^2-12r+52=0 are r=6+4i,r=6-4i
Find the zeros of r^2-12r+52=0
The zeros of r^2-12r+52=0 are r=6+4i,r=6-4i