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

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

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

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

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

y = 1 + 5, y = 1 - 5

Decimal Notation

y = 3.23606 \dots, y = - 1.23606 \dots

Solution steps

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

Multiply both sides by

2

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

Move

2

to the left side

y^{2} - 4 = 2 y

Move

2 y

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 4) = 25

y_{1, 2} = \frac{- ( - 2 ) \pm 2 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 2 ) + 2 5}{2 \cdot 1}, y_{2} = \frac{- ( - 2 ) - 2 5}{2 \cdot 1}

y = \frac{- ( - 2 ) + 2 5}{2 \cdot 1} : 1 + 5

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

The solutions to the quadratic equation are:

y = 1 + 5, y = 1 - 5

x^{2} + 1.3 = 0

3 x^{2} - 36 x + 108 = 0

x^{2} + 4 x - 6 = 0

(x + 4)^{2} = 2 x (5 x - 1) - 17 (x - 2)

9 x^{2} + 21 x + 10 = 0

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