What are the solutions to the equation solvefor x,a=(4x+y)^2 ?

The solutions to the equation solvefor x,a=(4x+y)^2 are x=(sqrt(a)-y)/4 ,x=(-sqrt(a)-y)/4

Find the zeros of solvefor x,a=(4x+y)^2

The zeros of solvefor x,a=(4x+y)^2 are x=(sqrt(a)-y)/4 ,x=(-sqrt(a)-y)/4

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solve for

x, a = (4 x + y)^{2}

x = \frac{a - y}{4}, x = \frac{- a - y}{4}

Solution steps

a = (4 x + y)^{2}

Switch sides

(4 x + y)^{2} = a

For

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

the solutions are

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

Solve

4 x + y = a : x = \frac{a - y}{4}

Solve

4 x + y = - a : x = \frac{- a - y}{4}

The solutions to the quadratic equation are:

x = \frac{a - y}{4}, x = \frac{- a - y}{4}

(x - 1) (x^{2} - x + 1) = x^{3} - 1

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

x, \frac{( x ^{2} + m ^{2} )}{v} = \frac{( v ^{2} - s )}{m}

x, x^{2} + y^{2} + z^{2} = x + y + z + 84

x, [3 p + 1] \cdot x^{2} + 2 [p + 1] \cdot x + p = 0

What are the solutions to the equation solvefor x,a=(4x+y)^2 ?
The solutions to the equation solvefor x,a=(4x+y)^2 are x=(sqrt(a)-y)/4 ,x=(-sqrt(a)-y)/4
Find the zeros of solvefor x,a=(4x+y)^2
The zeros of solvefor x,a=(4x+y)^2 are x=(sqrt(a)-y)/4 ,x=(-sqrt(a)-y)/4