What are the solutions to the equation t^2+3t-6=0 ?

The solutions to the equation t^2+3t-6=0 are t=(-3+sqrt(33))/2 ,t=(-3-sqrt(33))/2

Find the zeros of t^2+3t-6=0

The zeros of t^2+3t-6=0 are t=(-3+sqrt(33))/2 ,t=(-3-sqrt(33))/2

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t^{2} + 3 t - 6 = 0

t = \frac{- 3 + 33}{2}, t = \frac{- 3 - 33}{2}

Decimal Notation

t = 1.37228 \dots, t = - 4.37228 \dots

Solution steps

t^{2} + 3 t - 6 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 6) = 33

t_{1, 2} = \frac{- 3 \pm 33}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 3 + 33}{2 \cdot 1}, t_{2} = \frac{- 3 - 33}{2 \cdot 1}

t = \frac{- 3 + 33}{2 \cdot 1} : \frac{- 3 + 33}{2}

t = \frac{- 3 - 33}{2 \cdot 1} : \frac{- 3 - 33}{2}

The solutions to the quadratic equation are:

t = \frac{- 3 + 33}{2}, t = \frac{- 3 - 33}{2}

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a, a^{3} + b^{3} = (a + b)^{3}

\frac{4}{8} x^{2} + \frac{3}{2} x - 0.5 = 0

(x - 5) (x - 6) = \frac{25}{2 4 ^{2}}

46 \cdot x^{2} - 13 x - 26 = 0

What are the solutions to the equation t^2+3t-6=0 ?
The solutions to the equation t^2+3t-6=0 are t=(-3+sqrt(33))/2 ,t=(-3-sqrt(33))/2
Find the zeros of t^2+3t-6=0
The zeros of t^2+3t-6=0 are t=(-3+sqrt(33))/2 ,t=(-3-sqrt(33))/2