What are the solutions to the equation y^2-8y+22=0 ?

The solutions to the equation y^2-8y+22=0 are y=4+sqrt(6)i,y=4-sqrt(6)i

Find the zeros of y^2-8y+22=0

The zeros of y^2-8y+22=0 are y=4+sqrt(6)i,y=4-sqrt(6)i

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y^{2} - 8 y + 22 = 0

y = 4 + 6 i, y = 4 - 6 i

Solution steps

y^{2} - 8 y + 22 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 22}{2 \cdot 1}

Simplify

(- 8)^{2} - 4 \cdot 1 \cdot 22 : 26 i

y_{1, 2} = \frac{- ( - 8 ) \pm 2 6 i}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 2 6 i}{2 \cdot 1}, y_{2} = \frac{- ( - 8 ) - 2 6 i}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

y = 4 + 6 i, y = 4 - 6 i

(x + 3) (x - 2) = 1

x^{2} + 10 x = 3

4 x^{2} + 5 = 54

4 x^{2} + 5 = 41

4^{2} + b^{2} = 1 0^{2}

What are the solutions to the equation y^2-8y+22=0 ?
The solutions to the equation y^2-8y+22=0 are y=4+sqrt(6)i,y=4-sqrt(6)i
Find the zeros of y^2-8y+22=0
The zeros of y^2-8y+22=0 are y=4+sqrt(6)i,y=4-sqrt(6)i