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

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

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

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

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solve for

t, x = t^{2} + t

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

Solution steps

x = t^{2} + t

Switch sides

t^{2} + t = x

Move

x

to the left side

t^{2} + t - x = 0

Solve with the quadratic formula

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

Simplify

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

t_{1, 2} = \frac{- 1 \pm 1 + 4 x}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 1 + 1 + 4 x}{2 \cdot 1}, t_{2} = \frac{- 1 - 1 + 4 x}{2 \cdot 1}

t = \frac{- 1 + 1 + 4 x}{2 \cdot 1} : \frac{- 1 + 1 + 4 x}{2}

t = \frac{- 1 - 1 + 4 x}{2 \cdot 1} : \frac{- 1 - 1 + 4 x}{2}

The solutions to the quadratic equation are:

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

0 = - 6 c^{2} + 30 c + 300

1 7^{2} = x^{2} + 5^{2}

250 = 50 x - 300 - 0.01 x^{2}

5 x^{2} + 70 = 50 x

3 h^{2} - 5 h + 2 = 0

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