What are the solutions to the equation solvefor x,y=x^2-7x+12 ?

The solutions to the equation solvefor x,y=x^2-7x+12 are x=(7+sqrt(4y+1))/2 ,x=(7-sqrt(4y+1))/2

Find the zeros of solvefor x,y=x^2-7x+12

The zeros of solvefor x,y=x^2-7x+12 are x=(7+sqrt(4y+1))/2 ,x=(7-sqrt(4y+1))/2

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solve for

x, y = x^{2} - 7 x + 12

x = \frac{7 + 4 y + 1}{2}, x = \frac{7 - 4 y + 1}{2}

Solution steps

y = x^{2} - 7 x + 12

Switch sides

x^{2} - 7 x + 12 = y

Move

y

to the left side

x^{2} - 7 x + 12 - y = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot ( 12 - y )}{2 \cdot 1}

Simplify

(- 7)^{2} - 4 \cdot 1 \cdot (12 - y) : 4 y + 1

x_{1, 2} = \frac{- ( - 7 ) \pm 4 y + 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 4 y + 1}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 4 y + 1}{2 \cdot 1}

x = \frac{- ( - 7 ) + 4 y + 1}{2 \cdot 1} : \frac{7 + 4 y + 1}{2}

x = \frac{- ( - 7 ) - 4 y + 1}{2 \cdot 1} : \frac{7 - 4 y + 1}{2}

The solutions to the quadratic equation are:

x = \frac{7 + 4 y + 1}{2}, x = \frac{7 - 4 y + 1}{2}

2 x^{2} + x = 45

x^{2} - 4 x = 12

4 m^{2} + 9 = - 16

2 x^{2} + x = - 1

x, x^{2} - 4 x + 10 = 0

What are the solutions to the equation solvefor x,y=x^2-7x+12 ?
The solutions to the equation solvefor x,y=x^2-7x+12 are x=(7+sqrt(4y+1))/2 ,x=(7-sqrt(4y+1))/2
Find the zeros of solvefor x,y=x^2-7x+12
The zeros of solvefor x,y=x^2-7x+12 are x=(7+sqrt(4y+1))/2 ,x=(7-sqrt(4y+1))/2