What are the solutions to the equation solvefor y,h=y^2+y-6 ?

The solutions to the equation solvefor y,h=y^2+y-6 are y=(-1+sqrt(4h+25))/2 ,y=(-1-sqrt(4h+25))/2

Find the zeros of solvefor y,h=y^2+y-6

The zeros of solvefor y,h=y^2+y-6 are y=(-1+sqrt(4h+25))/2 ,y=(-1-sqrt(4h+25))/2

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y, h = y^{2} + y - 6

y = \frac{- 1 + 4 h + 25}{2}, y = \frac{- 1 - 4 h + 25}{2}

Solution steps

h = y^{2} + y - 6

Switch sides

y^{2} + y - 6 = h

Move

h

to the left side

y^{2} + y - 6 - h = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 6 - h )}{2 \cdot 1}

Simplify

1^{2} - 4 \cdot 1 \cdot (- 6 - h) : 4 h + 25

y_{1, 2} = \frac{- 1 \pm 4 h + 25}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 1 + 4 h + 25}{2 \cdot 1}, y_{2} = \frac{- 1 - 4 h + 25}{2 \cdot 1}

y = \frac{- 1 + 4 h + 25}{2 \cdot 1} : \frac{- 1 + 4 h + 25}{2}

y = \frac{- 1 - 4 h + 25}{2 \cdot 1} : \frac{- 1 - 4 h + 25}{2}

The solutions to the quadratic equation are:

y = \frac{- 1 + 4 h + 25}{2}, y = \frac{- 1 - 4 h + 25}{2}

1.1 \cdot 692.957 = 22.4 \cdot 2612 \cdot x \cdot (1900 - \frac{x}{2})

6 r^{2} + 2 r - 1 = 0

6 w^{2} - w = 5

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

0. 6^{2} + x^{2} = 1

What are the solutions to the equation solvefor y,h=y^2+y-6 ?
The solutions to the equation solvefor y,h=y^2+y-6 are y=(-1+sqrt(4h+25))/2 ,y=(-1-sqrt(4h+25))/2
Find the zeros of solvefor y,h=y^2+y-6
The zeros of solvefor y,h=y^2+y-6 are y=(-1+sqrt(4h+25))/2 ,y=(-1-sqrt(4h+25))/2