What are the solutions to the equation y^2-3=2y ?

The solutions to the equation y^2-3=2y are y=3,y=-1

Find the zeros of y^2-3=2y

The zeros of y^2-3=2y are y=3,y=-1

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y^{2} - 3 = 2 y

y = 3, y = - 1

Solution steps

y^{2} - 3 = 2 y

Move

2 y

to the left side

y^{2} - 3 - 2 y = 0

Write in the standard form

a x^{2} + b x + c = 0

y^{2} - 2 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 3) = 4

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

Separate the solutions

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

y = \frac{- ( - 2 ) + 4}{2 \cdot 1} : 3

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

The solutions to the quadratic equation are:

y = 3, y = - 1

x^{4} y^{2} + x y^{5}

5^{2} + 3^{2} = x^{2}

\frac{1}{3} (- 1)

6 \cdot 5 \cdot 4

- 8 n^{2} - 3 n + 2 = 8

What are the solutions to the equation y^2-3=2y ?
The solutions to the equation y^2-3=2y are y=3,y=-1
Find the zeros of y^2-3=2y
The zeros of y^2-3=2y are y=3,y=-1