What are the solutions to the equation y(y-4)=1 ?

The solutions to the equation y(y-4)=1 are y=2+sqrt(5),y=2-sqrt(5)

Find the zeros of y(y-4)=1

The zeros of y(y-4)=1 are y=2+sqrt(5),y=2-sqrt(5)

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y (y - 4) = 1

y = 2 + 5, y = 2 - 5

Decimal Notation

y = 4.23606 \dots, y = - 0.23606 \dots

Solution steps

y (y - 4) = 1

Expand

y (y - 4) : y^{2} - 4 y

y^{2} - 4 y = 1

Move

1

to the left side

y^{2} - 4 y - 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 1) = 25

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

Separate the solutions

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

y = \frac{- ( - 4 ) + 2 5}{2 \cdot 1} : 2 + 5

y = \frac{- ( - 4 ) - 2 5}{2 \cdot 1} : 2 - 5

The solutions to the quadratic equation are:

y = 2 + 5, y = 2 - 5

2 x^{3} - x^{2} (2 i - 1) - x (3 i + 5) + 2 + 2 i

7 m^{7} n - 28 m n^{3}

\frac{1 ^{2}}{2}

2 x^{2} + 3 x - 27 = 0

\frac{1 0 ^{0} r ^{11} s}{3 ^{2}}

What are the solutions to the equation y(y-4)=1 ?
The solutions to the equation y(y-4)=1 are y=2+sqrt(5),y=2-sqrt(5)
Find the zeros of y(y-4)=1
The zeros of y(y-4)=1 are y=2+sqrt(5),y=2-sqrt(5)