What are the solutions to the equation 2y^2-3y=1 ?

The solutions to the equation 2y^2-3y=1 are y=(3+sqrt(17))/4 ,y=(3-sqrt(17))/4

Find the zeros of 2y^2-3y=1

The zeros of 2y^2-3y=1 are y=(3+sqrt(17))/4 ,y=(3-sqrt(17))/4

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2 y^{2} - 3 y = 1

y = \frac{3 + 17}{4}, y = \frac{3 - 17}{4}

Decimal Notation

y = 1.78077 \dots, y = - 0.28077 \dots

Solution steps

2 y^{2} - 3 y = 1

Move

1

to the left side

2 y^{2} - 3 y - 1 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

y = \frac{- ( - 3 ) + 17}{2 \cdot 2} : \frac{3 + 17}{4}

y = \frac{- ( - 3 ) - 17}{2 \cdot 2} : \frac{3 - 17}{4}

The solutions to the quadratic equation are:

y = \frac{3 + 17}{4}, y = \frac{3 - 17}{4}

- 5 x^{2} + 70 x - 370 = 0

6 x^{2} - 3 x + 7 = 0

x, x^{2} + 9 y^{2} = k

\frac{x ^{2}}{4} = 16

x^{2} - 4 x - 50 = 0

What are the solutions to the equation 2y^2-3y=1 ?
The solutions to the equation 2y^2-3y=1 are y=(3+sqrt(17))/4 ,y=(3-sqrt(17))/4
Find the zeros of 2y^2-3y=1
The zeros of 2y^2-3y=1 are y=(3+sqrt(17))/4 ,y=(3-sqrt(17))/4