What are the solutions to the equation r^2-12r+40=0 ?

The solutions to the equation r^2-12r+40=0 are r=6+2i,r=6-2i

Find the zeros of r^2-12r+40=0

The zeros of r^2-12r+40=0 are r=6+2i,r=6-2i

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r^{2} - 12 r + 40 = 0

r = 6 + 2 i, r = 6 - 2 i

Solution steps

r^{2} - 12 r + 40 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 40}{2 \cdot 1}

Simplify

(- 12)^{2} - 4 \cdot 1 \cdot 40 : 4 i

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

Separate the solutions

r_{1} = \frac{- ( - 12 ) + 4 i}{2 \cdot 1}, r_{2} = \frac{- ( - 12 ) - 4 i}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

r = 6 + 2 i, r = 6 - 2 i

2 y^{2} - 9 y - 5

20 - 5 + 45

∣ x - 6 ∣ + 4 = 10

16 s^{4} - 4 t^{4}

12 \geq f

What are the solutions to the equation r^2-12r+40=0 ?
The solutions to the equation r^2-12r+40=0 are r=6+2i,r=6-2i
Find the zeros of r^2-12r+40=0
The zeros of r^2-12r+40=0 are r=6+2i,r=6-2i