What are the solutions to the equation t^2+3t-1=0 ?

The solutions to the equation t^2+3t-1=0 are t=(-3+sqrt(13))/2 ,t=(-3-sqrt(13))/2

Find the zeros of t^2+3t-1=0

The zeros of t^2+3t-1=0 are t=(-3+sqrt(13))/2 ,t=(-3-sqrt(13))/2

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t^{2} + 3 t - 1 = 0

t = \frac{- 3 + 13}{2}, t = \frac{- 3 - 13}{2}

Decimal Notation

t = 0.30277 \dots, t = - 3.30277 \dots

Solution steps

t^{2} + 3 t - 1 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 1) = 13

t_{1, 2} = \frac{- 3 \pm 13}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 3 + 13}{2 \cdot 1}, t_{2} = \frac{- 3 - 13}{2 \cdot 1}

t = \frac{- 3 + 13}{2 \cdot 1} : \frac{- 3 + 13}{2}

t = \frac{- 3 - 13}{2 \cdot 1} : \frac{- 3 - 13}{2}

The solutions to the quadratic equation are:

t = \frac{- 3 + 13}{2}, t = \frac{- 3 - 13}{2}

\frac{2}{3} + \frac{2}{6}

9 x^{2} - 24 x y + 16 y^{2}

2 x^{2} - 7 = - 135

(x - 6)^{2} (28 - 4 x) > 0

2 x + 13 < 3

What are the solutions to the equation t^2+3t-1=0 ?
The solutions to the equation t^2+3t-1=0 are t=(-3+sqrt(13))/2 ,t=(-3-sqrt(13))/2
Find the zeros of t^2+3t-1=0
The zeros of t^2+3t-1=0 are t=(-3+sqrt(13))/2 ,t=(-3-sqrt(13))/2