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

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

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

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

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y + 1 = \frac{y ^{2} - 6}{2}

y = 4, y = - 2

Solution steps

y + 1 = \frac{y ^{2} - 6}{2}

Multiply both sides by

2

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

Switch sides

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

Move

2

to the left side

y^{2} - 8 = 2 y

Move

2 y

to the left side

y^{2} - 8 - 2 y = 0

Write in the standard form

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

y^{2} - 2 y - 8 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 8) = 6

y_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 1}, y_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 1}

y = \frac{- ( - 2 ) + 6}{2 \cdot 1} : 4

y = \frac{- ( - 2 ) - 6}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

y = 4, y = - 2

(1 - x)^{2} - 2 = 0

x^{2} - 49 = (x + 7)

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

n^{2} + 14 n - 5 = 0

x^{2} = 10 x - 20

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