What are the solutions to the equation y^2-4y+40=0 ?

The solutions to the equation y^2-4y+40=0 are y=2+6i,y=2-6i

Find the zeros of y^2-4y+40=0

The zeros of y^2-4y+40=0 are y=2+6i,y=2-6i

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y^{2} - 4 y + 40 = 0

y = 2 + 6 i, y = 2 - 6 i

Solution steps

y^{2} - 4 y + 40 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 2 + 6 i, y = 2 - 6 i

21 x - 100 = x^{2} + 21 - x

- x^{2} + 9 x + 10 = 0

x^{2} + 8 x + 8 = \frac{( 0 ) - ( 0 )}{( 4 ) - ( - 3 )}

(- 1 - x) (1 - x) = 0

x, 9 x^{2} + 4 y^{2} + 36 x - 24 y + 216 = 0

What are the solutions to the equation y^2-4y+40=0 ?
The solutions to the equation y^2-4y+40=0 are y=2+6i,y=2-6i
Find the zeros of y^2-4y+40=0
The zeros of y^2-4y+40=0 are y=2+6i,y=2-6i