What are the solutions to the equation solvefor x,f=(2x-3y)^2 ?

The solutions to the equation solvefor x,f=(2x-3y)^2 are x=(sqrt(f)+3y)/2 ,x=(-sqrt(f)+3y)/2

Find the zeros of solvefor x,f=(2x-3y)^2

The zeros of solvefor x,f=(2x-3y)^2 are x=(sqrt(f)+3y)/2 ,x=(-sqrt(f)+3y)/2

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solve for

x, f = (2 x - 3 y)^{2}

x = \frac{f + 3 y}{2}, x = \frac{- f + 3 y}{2}

Solution steps

f = (2 x - 3 y)^{2}

Switch sides

(2 x - 3 y)^{2} = f

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

2 x - 3 y = f : x = \frac{f + 3 y}{2}

Solve

2 x - 3 y = - f : x = \frac{- f + 3 y}{2}

The solutions to the quadratic equation are:

x = \frac{f + 3 y}{2}, x = \frac{- f + 3 y}{2}

x (x + 8) = 9

7 a^{2} + 2 a = 5

x^{2} + 12 x + 35 = 0

(4 - x) (- 1 - x) - 6 = 0

2 x^{2} - 12 x - 224 = 0

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