What are the solutions to the equation solvefor t,h=t^2+2t ?

The solutions to the equation solvefor t,h=t^2+2t are t=-1+sqrt(h+1),t=-sqrt(h+1)-1

Find the zeros of solvefor t,h=t^2+2t

The zeros of solvefor t,h=t^2+2t are t=-1+sqrt(h+1),t=-sqrt(h+1)-1

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solve for

t, h = t^{2} + 2 t

t = - 1 + h + 1, t = - h + 1 - 1

Solution steps

h = t^{2} + 2 t

Switch sides

t^{2} + 2 t = h

Move

h

to the left side

t^{2} + 2 t - h = 0

Solve with the quadratic formula

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

Simplify

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

t_{1, 2} = \frac{- 2 \pm 2 1 + h}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 2 + 2 1 + h}{2 \cdot 1}, t_{2} = \frac{- 2 - 2 1 + h}{2 \cdot 1}

t = \frac{- 2 + 2 1 + h}{2 \cdot 1} : - 1 + h + 1

t = \frac{- 2 - 2 1 + h}{2 \cdot 1} : - h + 1 - 1

The solutions to the quadratic equation are:

t = - 1 + h + 1, t = - h + 1 - 1

2 x^{2} - 9 = 12 x

v^{2} - 3 = - 16

X^{2} - 6 X + 2 = 0

e^{2} = 90

9 x^{2} - 2260 x + 129600 = 0

What are the solutions to the equation solvefor t,h=t^2+2t ?
The solutions to the equation solvefor t,h=t^2+2t are t=-1+sqrt(h+1),t=-sqrt(h+1)-1
Find the zeros of solvefor t,h=t^2+2t
The zeros of solvefor t,h=t^2+2t are t=-1+sqrt(h+1),t=-sqrt(h+1)-1