What are the solutions to the equation solvefor t,x=t^2+3t ?

The solutions to the equation solvefor t,x=t^2+3t are t=(-3+sqrt(9+4x))/2 ,t=(-3-sqrt(9+4x))/2

Find the zeros of solvefor t,x=t^2+3t

The zeros of solvefor t,x=t^2+3t are t=(-3+sqrt(9+4x))/2 ,t=(-3-sqrt(9+4x))/2

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solve for

t, x = t^{2} + 3 t

t = \frac{- 3 + 9 + 4 x}{2}, t = \frac{- 3 - 9 + 4 x}{2}

Solution steps

x = t^{2} + 3 t

Switch sides

t^{2} + 3 t = x

Move

x

to the left side

t^{2} + 3 t - x = 0

Solve with the quadratic formula

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

Simplify

3^{2} - 4 \cdot 1 \cdot (- x) : 9 + 4 x

t_{1, 2} = \frac{- 3 \pm 9 + 4 x}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 3 + 9 + 4 x}{2 \cdot 1}, t_{2} = \frac{- 3 - 9 + 4 x}{2 \cdot 1}

t = \frac{- 3 + 9 + 4 x}{2 \cdot 1} : \frac{- 3 + 9 + 4 x}{2}

t = \frac{- 3 - 9 + 4 x}{2 \cdot 1} : \frac{- 3 - 9 + 4 x}{2}

The solutions to the quadratic equation are:

t = \frac{- 3 + 9 + 4 x}{2}, t = \frac{- 3 - 9 + 4 x}{2}

4 u^{2} = - 5 u + 6

4 x^{2} - 9 x - 6 = 0

u^{2} - 6 u

2 - x^{2} = 2 - x

\frac{x ^{2} - 1}{4} - \frac{3 x - 1}{6} = \frac{x}{2}

What are the solutions to the equation solvefor t,x=t^2+3t ?
The solutions to the equation solvefor t,x=t^2+3t are t=(-3+sqrt(9+4x))/2 ,t=(-3-sqrt(9+4x))/2
Find the zeros of solvefor t,x=t^2+3t
The zeros of solvefor t,x=t^2+3t are t=(-3+sqrt(9+4x))/2 ,t=(-3-sqrt(9+4x))/2