What are the solutions to the equation solvefor X,Y=X^2-4X+3 ?

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

Find the zeros of solvefor X,Y=X^2-4X+3

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

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solve for

X, Y = X^{2} - 4 X + 3

X = 2 + Y + 1, X = 2 - Y + 1

Solution steps

Y = X^{2} - 4 X + 3

Switch sides

X^{2} - 4 X + 3 = Y

Move

Y

to the left side

X^{2} - 4 X + 3 - Y = 0

Solve with the quadratic formula

X_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( 3 - Y )}{2 \cdot 1}

Simplify

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

X_{1, 2} = \frac{- ( - 4 ) \pm 2 Y + 1}{2 \cdot 1}

Separate the solutions

X_{1} = \frac{- ( - 4 ) + 2 Y + 1}{2 \cdot 1}, X_{2} = \frac{- ( - 4 ) - 2 Y + 1}{2 \cdot 1}

X = \frac{- ( - 4 ) + 2 Y + 1}{2 \cdot 1} : 2 + Y + 1

X = \frac{- ( - 4 ) - 2 Y + 1}{2 \cdot 1} : 2 - Y + 1

The solutions to the quadratic equation are:

X = 2 + Y + 1, X = 2 - Y + 1

- 7 x^{2} + 7 x + 140 = 0

x, x^{2} + y^{2} - 2 y - z + 1 = 0

x, x^{2} + 4 x + 8 = 0

(x - 4)^{2} = 8

12 x + 15 = 3 x^{2}

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