What are the solutions to the equation (2y-1)^2-4=6y-1 ?

The solutions to the equation (2y-1)^2-4=6y-1 are y=(5+sqrt(33))/4 ,y=(5-sqrt(33))/4

Find the zeros of (2y-1)^2-4=6y-1

The zeros of (2y-1)^2-4=6y-1 are y=(5+sqrt(33))/4 ,y=(5-sqrt(33))/4

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(2 y - 1)^{2} - 4 = 6 y - 1

y = \frac{5 + 33}{4}, y = \frac{5 - 33}{4}

Decimal

y = 2.68614 \dots, y = - 0.18614 \dots

Solution steps

(2 y - 1)^{2} - 4 = 6 y - 1

Expand

(2 y - 1)^{2} - 4 : 4 y^{2} - 4 y - 3

4 y^{2} - 4 y - 3 = 6 y - 1

Move

1

to the left side

4 y^{2} - 4 y - 2 = 6 y

Move

6 y

to the left side

4 y^{2} - 10 y - 2 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 4 ( - 2 )}{2 \cdot 4}

(- 10)^{2} - 4 \cdot 4 (- 2) = 233

y_{1, 2} = \frac{- ( - 10 ) \pm 2 33}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- ( - 10 ) + 2 33}{2 \cdot 4}, y_{2} = \frac{- ( - 10 ) - 2 33}{2 \cdot 4}

y = \frac{- ( - 10 ) + 2 33}{2 \cdot 4} : \frac{5 + 33}{4}

y = \frac{- ( - 10 ) - 2 33}{2 \cdot 4} : \frac{5 - 33}{4}

The solutions to the quadratic equation are:

y = \frac{5 + 33}{4}, y = \frac{5 - 33}{4}

- 33 p = - p^{2}

2 x^{2} - 12 x - 9 = 0

4 q^{2} + q + 1 = - 3 q

0 = 2 x^{2} - 8 x

10 n^{2} + 15 n - 25 = 6 n^{2}

What are the solutions to the equation (2y-1)^2-4=6y-1 ?
The solutions to the equation (2y-1)^2-4=6y-1 are y=(5+sqrt(33))/4 ,y=(5-sqrt(33))/4
Find the zeros of (2y-1)^2-4=6y-1
The zeros of (2y-1)^2-4=6y-1 are y=(5+sqrt(33))/4 ,y=(5-sqrt(33))/4