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

The solutions to the equation solvefor x,y=x^2+2x+4 are x=-1+sqrt(y-3),x=-sqrt(y-3)-1

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

The zeros of solvefor x,y=x^2+2x+4 are x=-1+sqrt(y-3),x=-sqrt(y-3)-1

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solve for

x, y = x^{2} + 2 x + 4

x = - 1 + y - 3, x = - y - 3 - 1

Solution steps

y = x^{2} + 2 x + 4

Switch sides

x^{2} + 2 x + 4 = y

Move

y

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( 4 - y )}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot (4 - y) : 2 y - 3

x_{1, 2} = \frac{- 2 \pm 2 y - 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 2 y - 3}{2 \cdot 1}, x_{2} = \frac{- 2 - 2 y - 3}{2 \cdot 1}

x = \frac{- 2 + 2 y - 3}{2 \cdot 1} : - 1 + y - 3

x = \frac{- 2 - 2 y - 3}{2 \cdot 1} : - y - 3 - 1

The solutions to the quadratic equation are:

x = - 1 + y - 3, x = - y - 3 - 1

26 x^{2} + 138 x + 213 = 0

9 x^{2} + 240 x - 4800 = 0

- x^{2} - x + 6 = 0

x^{2} + 12 x + \frac{1}{3} = 0

y^{2} = 18 y + 18

What are the solutions to the equation solvefor x,y=x^2+2x+4 ?
The solutions to the equation solvefor x,y=x^2+2x+4 are x=-1+sqrt(y-3),x=-sqrt(y-3)-1
Find the zeros of solvefor x,y=x^2+2x+4
The zeros of solvefor x,y=x^2+2x+4 are x=-1+sqrt(y-3),x=-sqrt(y-3)-1