What are the solutions to the equation y^2-4y=-5 ?

The solutions to the equation y^2-4y=-5 are y=2+i,y=2-i

Find the zeros of y^2-4y=-5

The zeros of y^2-4y=-5 are y=2+i,y=2-i

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y^{2} - 4 y = - 5

y = 2 + i, y = 2 - i

Solution steps

y^{2} - 4 y = - 5

Move

5

to the left side

y^{2} - 4 y + 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 5 : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 2 + i, y = 2 - i

2 x^{2} = 8 x + 4

2 x^{2} = 8 x - 9

16 x^{2} - 4 x = 0

x^{2} - 2 = 3 x

55 = - 16 t^{2} + 90 t

What are the solutions to the equation y^2-4y=-5 ?
The solutions to the equation y^2-4y=-5 are y=2+i,y=2-i
Find the zeros of y^2-4y=-5
The zeros of y^2-4y=-5 are y=2+i,y=2-i