What are the solutions to the equation solvefor x,ln(y)=(x^2)/2+c ?

The solutions to the equation solvefor x,ln(y)=(x^2)/2+c are x=sqrt(2ln(y)-2c),x=-sqrt(2ln(y)-2c)

Find the zeros of solvefor x,ln(y)=(x^2)/2+c

The zeros of solvefor x,ln(y)=(x^2)/2+c are x=sqrt(2ln(y)-2c),x=-sqrt(2ln(y)-2c)

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solve for

x, ln (y) = \frac{x ^{2}}{2} + c

x = 2 ln (y) - 2 c, x = - 2 ln (y) - 2 c

Solution steps

ln (y) = \frac{x ^{2}}{2} + c

Switch sides

\frac{x ^{2}}{2} + c = ln (y)

Move

c

to the right side

\frac{x ^{2}}{2} = ln (y) - c

Multiply both sides by

2

x^{2} = 2 ln (y) - 2 c

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 2 ln (y) - 2 c, x = - 2 ln (y) - 2 c

x^{2} + 20 x - 400 = 0

10 x^{2} - 33 x + 20 = 0

4 x^{2} - 36 x + 81

1.36 x^{2} + 27.8 x + 304 = 500

(400 + j^{1000}) I^{2} - j^{90} I^{1} = 0

What are the solutions to the equation solvefor x,ln(y)=(x^2)/2+c ?
The solutions to the equation solvefor x,ln(y)=(x^2)/2+c are x=sqrt(2ln(y)-2c),x=-sqrt(2ln(y)-2c)
Find the zeros of solvefor x,ln(y)=(x^2)/2+c
The zeros of solvefor x,ln(y)=(x^2)/2+c are x=sqrt(2ln(y)-2c),x=-sqrt(2ln(y)-2c)