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

The solutions to the equation y^2-3y=2 are y=(3+sqrt(17))/2 ,y=(3-sqrt(17))/2

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

The zeros of y^2-3y=2 are y=(3+sqrt(17))/2 ,y=(3-sqrt(17))/2

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y^{2} - 3 y = 2

y = \frac{3 + 17}{2}, y = \frac{3 - 17}{2}

Decimal Notation

y = 3.56155 \dots, y = - 0.56155 \dots

Solution steps

y^{2} - 3 y = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 2) = 17

y_{1, 2} = \frac{- ( - 3 ) \pm 17}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 17}{2 \cdot 1}, y_{2} = \frac{- ( - 3 ) - 17}{2 \cdot 1}

y = \frac{- ( - 3 ) + 17}{2 \cdot 1} : \frac{3 + 17}{2}

y = \frac{- ( - 3 ) - 17}{2 \cdot 1} : \frac{3 - 17}{2}

The solutions to the quadratic equation are:

y = \frac{3 + 17}{2}, y = \frac{3 - 17}{2}

- x^{2} + 3 = x^{2} + 1

16 x^{2} + 9 x - 12 = 4 x + 9 x^{2}

2 x^{2} + 2 x = 24

x, x^{2} y^{2} - 18 x = 3

- 16 t^{2} + 160 t + 120 = 0

What are the solutions to the equation y^2-3y=2 ?
The solutions to the equation y^2-3y=2 are y=(3+sqrt(17))/2 ,y=(3-sqrt(17))/2
Find the zeros of y^2-3y=2
The zeros of y^2-3y=2 are y=(3+sqrt(17))/2 ,y=(3-sqrt(17))/2