What are the solutions to the equation solvefor y,x=y^2-8y+3 ?

The solutions to the equation solvefor y,x=y^2-8y+3 are y=4+sqrt(x+13),y=4-sqrt(x+13)

Find the zeros of solvefor y,x=y^2-8y+3

The zeros of solvefor y,x=y^2-8y+3 are y=4+sqrt(x+13),y=4-sqrt(x+13)

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solve for

y, x = y^{2} - 8 y + 3

y = 4 + x + 13, y = 4 - x + 13

Solution steps

x = y^{2} - 8 y + 3

Switch sides

y^{2} - 8 y + 3 = x

Move

x

to the left side

y^{2} - 8 y + 3 - x = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( 3 - x )}{2 \cdot 1}

Simplify

(- 8)^{2} - 4 \cdot 1 \cdot (3 - x) : 2 x + 13

y_{1, 2} = \frac{- ( - 8 ) \pm 2 x + 13}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 2 x + 13}{2 \cdot 1}, y_{2} = \frac{- ( - 8 ) - 2 x + 13}{2 \cdot 1}

y = \frac{- ( - 8 ) + 2 x + 13}{2 \cdot 1} : 4 + x + 13

y = \frac{- ( - 8 ) - 2 x + 13}{2 \cdot 1} : 4 - x + 13

The solutions to the quadratic equation are:

y = 4 + x + 13, y = 4 - x + 13

0.05 x^{2} - 0.08 x = 0

- 2 x^{2} + 100 x

x^{2} - 40 x + 1200 = 0

(2 x + 4) (x - 7) = 2 (5 x - 3)

x, 0 = 2 x^{2} + y^{2}

What are the solutions to the equation solvefor y,x=y^2-8y+3 ?
The solutions to the equation solvefor y,x=y^2-8y+3 are y=4+sqrt(x+13),y=4-sqrt(x+13)
Find the zeros of solvefor y,x=y^2-8y+3
The zeros of solvefor y,x=y^2-8y+3 are y=4+sqrt(x+13),y=4-sqrt(x+13)