What are the solutions to the equation solvefor y,M=(y-3)^2+4y ?

The solutions to the equation solvefor y,M=(y-3)^2+4y are y=1+sqrt(M-8),y=1-sqrt(M-8)

Find the zeros of solvefor y,M=(y-3)^2+4y

The zeros of solvefor y,M=(y-3)^2+4y are y=1+sqrt(M-8),y=1-sqrt(M-8)

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solve for

y, M = (y - 3)^{2} + 4 y

y = 1 + M - 8, y = 1 - M - 8

Solution steps

M = (y - 3)^{2} + 4 y

Expand

(y - 3)^{2} + 4 y : y^{2} - 2 y + 9

M = y^{2} - 2 y + 9

Switch sides

y^{2} - 2 y + 9 = M

Move

M

to the left side

y^{2} - 2 y + 9 - M = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( 9 - M )}{2 \cdot 1}

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot (9 - M) : 2 M - 8

y_{1, 2} = \frac{- ( - 2 ) \pm 2 M - 8}{2 \cdot 1}

Separate the solutions

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

y = \frac{- ( - 2 ) + 2 M - 8}{2 \cdot 1} : 1 + M - 8

y = \frac{- ( - 2 ) - 2 M - 8}{2 \cdot 1} : 1 - M - 8

The solutions to the quadratic equation are:

y = 1 + M - 8, y = 1 - M - 8

\frac{x ^{2} - 5}{2} + \frac{2 x ^{2} - 3 x + 7}{2} = 5

s^{2} - s + 10 = 0

- 2 x^{2} + 28 x - 148 = 0

- 4 x^{2} + 12 x = 0

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

What are the solutions to the equation solvefor y,M=(y-3)^2+4y ?
The solutions to the equation solvefor y,M=(y-3)^2+4y are y=1+sqrt(M-8),y=1-sqrt(M-8)
Find the zeros of solvefor y,M=(y-3)^2+4y
The zeros of solvefor y,M=(y-3)^2+4y are y=1+sqrt(M-8),y=1-sqrt(M-8)