What is the answer to solvefor x,sqrt(x^2+x+1)=z-4 ?

The answer to solvefor x,sqrt(x^2+x+1)=z-4 is

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solvefor x,sqrt(x^2+x+1)=z-4

solve for

x, x^{2} + x + 1 = z - 4

x = \frac{- 1 + 4 z ^{2} - 32 z + 61}{2} {z \geq \frac{8 + 3}{2}}, x = \frac{- 1 - 4 z ^{2} - 32 z + 61}{2} {z \geq \frac{8 + 3}{2}}

Solution steps

x^{2} + x + 1 = z - 4

Square both sides:

x^{2} + x + 1 = z^{2} - 8 z + 16

x^{2} + x + 1 = z^{2} - 8 z + 16

Solve

x^{2} + x + 1 = z^{2} - 8 z + 16 : x = \frac{- 1 + 4 z ^{2} - 32 z + 61}{2}, x = \frac{- 1 - 4 z ^{2} - 32 z + 61}{2}

x = \frac{- 1 + 4 z ^{2} - 32 z + 61}{2}, x = \frac{- 1 - 4 z ^{2} - 32 z + 61}{2}

Verify Solutions:

x = \frac{- 1 + 4 z ^{2} - 32 z + 61}{2} {z \geq \frac{8 + 3}{2}}, x = \frac{- 1 - 4 z ^{2} - 32 z + 61}{2} {z \geq \frac{8 + 3}{2}}

The solutions are

x = \frac{- 1 + 4 z ^{2} - 32 z + 61}{2} {z \geq \frac{8 + 3}{2}}, x = \frac{- 1 - 4 z ^{2} - 32 z + 61}{2} {z \geq \frac{8 + 3}{2}}

y, \frac{s}{t} = v^{2} - y^{2}

x + 4 - x^{2} + 5 x + 4 = 0

3 x^{2} + 2 5 x - 5 = 0

c^{7.2} = 36

x, x^{\frac{1}{2}} + y^{\frac{1}{2}} = a^{\frac{1}{2}}

What is the answer to solvefor x,sqrt(x^2+x+1)=z-4 ?
The answer to solvefor x,sqrt(x^2+x+1)=z-4 is