What are the solutions to the equation 12+y^2-6y=0 ?

The solutions to the equation 12+y^2-6y=0 are y=3+sqrt(3)i,y=3-sqrt(3)i

Find the zeros of 12+y^2-6y=0

The zeros of 12+y^2-6y=0 are y=3+sqrt(3)i,y=3-sqrt(3)i

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12 + y^{2} - 6 y = 0

y = 3 + 3 i, y = 3 - 3 i

Solution steps

12 + y^{2} - 6 y = 0

Write in the standard form

a x^{2} + b x + c = 0

y^{2} - 6 y + 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

Simplify

(- 6)^{2} - 4 \cdot 1 \cdot 12 : 23 i

y_{1, 2} = \frac{- ( - 6 ) \pm 2 3 i}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 6 ) + 2 3 i}{2 \cdot 1}, y_{2} = \frac{- ( - 6 ) - 2 3 i}{2 \cdot 1}

y = \frac{- ( - 6 ) + 2 3 i}{2 \cdot 1} : 3 + 3 i

y = \frac{- ( - 6 ) - 2 3 i}{2 \cdot 1} : 3 - 3 i

The solutions to the quadratic equation are:

y = 3 + 3 i, y = 3 - 3 i

2 x^{2} - x - 1 = 0

5 x^{2} - 24 x + 16 = 0

3 v^{2} = v + 10

3 x^{2} + 5 x - 4 = x^{2} - 2 x

x^{2} = 16 x - 56

What are the solutions to the equation 12+y^2-6y=0 ?
The solutions to the equation 12+y^2-6y=0 are y=3+sqrt(3)i,y=3-sqrt(3)i
Find the zeros of 12+y^2-6y=0
The zeros of 12+y^2-6y=0 are y=3+sqrt(3)i,y=3-sqrt(3)i