What are the solutions to the equation y^2-8y+17=0 ?

The solutions to the equation y^2-8y+17=0 are y=4+i,y=4-i

Find the zeros of y^2-8y+17=0

The zeros of y^2-8y+17=0 are y=4+i,y=4-i

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y^{2} - 8 y + 17 = 0

y = 4 + i, y = 4 - i

Solution steps

y^{2} - 8 y + 17 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 17}{2 \cdot 1}

Simplify

(- 8)^{2} - 4 \cdot 1 \cdot 17 : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 4 + i, y = 4 - i

x^{2} = 4 x - 20

ln (36)

5 x = 6 x^{2} - 3

60 (1 0^{6}) = \frac{( - 900 + x ) ( 0.025 )}{( \frac{π}{2} ) ( 0.02 5 ^{4} )}

(- 1 + \frac{3}{4} i)^{4}

What are the solutions to the equation y^2-8y+17=0 ?
The solutions to the equation y^2-8y+17=0 are y=4+i,y=4-i
Find the zeros of y^2-8y+17=0
The zeros of y^2-8y+17=0 are y=4+i,y=4-i