What is the answer to solvefor x,r+s-2t=(2x+y)^{1/2} ?

The answer to solvefor x,r+s-2t=(2x+y)^{1/2} is x=(2rs-4rt+r^2+s^2-4st+4t^2-y)/2

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solvefor x,r+s-2t=(2x+y)^{1/2}

solve for

x, r + s - 2 t = (2 x + y)^{\frac{1}{2}}

x = \frac{2 rs - 4 r t + r ^{2} + s ^{2} - 4 s t + 4 t ^{2} - y}{2}

Solution steps

r + s - 2 t = (2 x + y)^{\frac{1}{2}}

Square both sides:

2 rs - 4 r t + r^{2} + s^{2} - 4 s t + 4 t^{2} = 2 x + y

2 rs - 4 r t + r^{2} + s^{2} - 4 s t + 4 t^{2} = 2 x + y

Solve

2 rs - 4 r t + r^{2} + s^{2} - 4 s t + 4 t^{2} = 2 x + y : x = \frac{2 rs - 4 r t + r ^{2} + s ^{2} - 4 s t + 4 t ^{2} - y}{2}

x = \frac{2 rs - 4 r t + r ^{2} + s ^{2} - 4 s t + 4 t ^{2} - y}{2}

x + 4 + 3 = 5 x + 11

x - 6 = 10

4 x + 1 + x + 2 = 1

t^{0.4} = 2 + 25 t

(q) = 20 + 2 q - q^{2} - 80

What is the answer to solvefor x,r+s-2t=(2x+y)^{1/2} ?
The answer to solvefor x,r+s-2t=(2x+y)^{1/2} is x=(2rs-4rt+r^2+s^2-4st+4t^2-y)/2