What are the solutions to the equation solvefor x,f=x^2-2x+3k ?

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

Find the zeros of solvefor x,f=x^2-2x+3k

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

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solve for

x, f = x^{2} - 2 x + 3 k

x = 1 + f + 1 - 3 k, x = 1 - f + 1 - 3 k

Solution steps

f = x^{2} - 2 x + 3 k

Switch sides

x^{2} - 2 x + 3 k = f

Move

f

to the left side

x^{2} - 2 x + 3 k - f = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( 3 k - f )}{2 \cdot 1}

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot (3 k - f) : 2 1 + f - 3 k

x_{1, 2} = \frac{- ( - 2 ) \pm 2 1 + f - 3 k}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 1 + f - 3 k}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 1 + f - 3 k}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 1 + f - 3 k}{2 \cdot 1} : 1 + f + 1 - 3 k

x = \frac{- ( - 2 ) - 2 1 + f - 3 k}{2 \cdot 1} : 1 - f + 1 - 3 k

The solutions to the quadratic equation are:

x = 1 + f + 1 - 3 k, x = 1 - f + 1 - 3 k

4 y^{2} + 9 = 12 y

a^{2} - 6 a + 8 = 0

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

- 4 x^{2} + 8000 x = 0

x^{2} + \frac{6540}{14500} x + \frac{34000}{14500} = 0

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