What are the solutions to the equation (y^2-3)/2+y/4 =1 ?

The solutions to the equation (y^2-3)/2+y/4 =1 are y=2,y=-5/2

Find the zeros of (y^2-3)/2+y/4 =1

The zeros of (y^2-3)/2+y/4 =1 are y=2,y=-5/2

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\frac{y ^{2} - 3}{2} + \frac{y}{4} = 1

y = 2, y = - \frac{5}{2}

Decimal Notation

y = 2, y = - 2.5

Solution steps

\frac{y ^{2} - 3}{2} + \frac{y}{4} = 1

Find Least Common Multiplier of

2, 4 : 4

Multiply by LCM=

4

\frac{y ^{2} - 3}{2} \cdot 4 + \frac{y}{4} \cdot 4 = 1 \cdot 4

Simplify

2 (y^{2} - 3) + y = 4

Expand

2 (y^{2} - 3) + y : 2 y^{2} - 6 + y

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

Move

4

to the left side

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 2 (- 10) = 9

y_{1, 2} = \frac{- 1 \pm 9}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 1 + 9}{2 \cdot 2}, y_{2} = \frac{- 1 - 9}{2 \cdot 2}

y = \frac{- 1 + 9}{2 \cdot 2} : 2

y = \frac{- 1 - 9}{2 \cdot 2} : - \frac{5}{2}

The solutions to the quadratic equation are:

y = 2, y = - \frac{5}{2}

n^{2} + 15 n + 43 = 0

x^{2} = 4 x + 8

B, [(A^{c} \cup B) - A] \cap (A \cup B) = B - A

x^{2} + 1 = 10

3 x^{2} + 15 x + 20 = 0

What are the solutions to the equation (y^2-3)/2+y/4 =1 ?
The solutions to the equation (y^2-3)/2+y/4 =1 are y=2,y=-5/2
Find the zeros of (y^2-3)/2+y/4 =1
The zeros of (y^2-3)/2+y/4 =1 are y=2,y=-5/2