What are the solutions to the equation y^2-4y+13=0 ?

The solutions to the equation y^2-4y+13=0 are y=2+3i,y=2-3i

Find the zeros of y^2-4y+13=0

The zeros of y^2-4y+13=0 are y=2+3i,y=2-3i

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y^{2} - 4 y + 13 = 0

y = 2 + 3 i, y = 2 - 3 i

Solution steps

y^{2} - 4 y + 13 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 13 : 6 i

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

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 6 i}{2 \cdot 1}, y_{2} = \frac{- ( - 4 ) - 6 i}{2 \cdot 1}

y = \frac{- ( - 4 ) + 6 i}{2 \cdot 1} : 2 + 3 i

y = \frac{- ( - 4 ) - 6 i}{2 \cdot 1} : 2 - 3 i

The solutions to the quadratic equation are:

y = 2 + 3 i, y = 2 - 3 i

\frac{a}{4} - \frac{5}{6} = - \frac{1}{2}

3 x^{2} - x = \frac{23}{2}

lo g_{2} (\frac{1}{x}) + lo g_{2} (\frac{1}{x ^{2}})

- 8 x - 27 < 5

x^{4} y - 2 x^{2} y + y

What are the solutions to the equation y^2-4y+13=0 ?
The solutions to the equation y^2-4y+13=0 are y=2+3i,y=2-3i
Find the zeros of y^2-4y+13=0
The zeros of y^2-4y+13=0 are y=2+3i,y=2-3i