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

The solutions to the equation solvefor x,f=x^2+4x+k are x=-2+sqrt(f+4-k),x=-sqrt(f+4-k)-2

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

The zeros of solvefor x,f=x^2+4x+k are x=-2+sqrt(f+4-k),x=-sqrt(f+4-k)-2

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solve for

x, f = x^{2} + 4 x + k

x = - 2 + f + 4 - k, x = - f + 4 - k - 2

Solution steps

f = x^{2} + 4 x + k

Switch sides

x^{2} + 4 x + k = f

Move

f

to the left side

x^{2} + 4 x + k - f = 0

Solve with the quadratic formula

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

Simplify

4^{2} - 4 \cdot 1 \cdot (k - f) : 2 4 + f - k

x_{1, 2} = \frac{- 4 \pm 2 4 + f - k}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 2 4 + f - k}{2 \cdot 1}, x_{2} = \frac{- 4 - 2 4 + f - k}{2 \cdot 1}

x = \frac{- 4 + 2 4 + f - k}{2 \cdot 1} : - 2 + f + 4 - k

x = \frac{- 4 - 2 4 + f - k}{2 \cdot 1} : - f + 4 - k - 2

The solutions to the quadratic equation are:

x = - 2 + f + 4 - k, x = - f + 4 - k - 2

20 x^{2} - 9 x = 20

(a + 5) x^{2} + (a + 3) x + 7 - 3 a = 0

x^{2} - 120 x + 3600 = 0

x^{2} - 3 x = 2 x + 14

23 = x^{2} + 7

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