What are the solutions to the equation t^2+5t-2=0 ?

The solutions to the equation t^2+5t-2=0 are t=(-5+sqrt(33))/2 ,t=(-5-sqrt(33))/2

Find the zeros of t^2+5t-2=0

The zeros of t^2+5t-2=0 are t=(-5+sqrt(33))/2 ,t=(-5-sqrt(33))/2

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t^{2} + 5 t - 2 = 0

t = \frac{- 5 + 33}{2}, t = \frac{- 5 - 33}{2}

Decimal Notation

t = 0.37228 \dots, t = - 5.37228 \dots

Solution steps

t^{2} + 5 t - 2 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 2) = 33

t_{1, 2} = \frac{- 5 \pm 33}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 5 + 33}{2 \cdot 1}, t_{2} = \frac{- 5 - 33}{2 \cdot 1}

t = \frac{- 5 + 33}{2 \cdot 1} : \frac{- 5 + 33}{2}

t = \frac{- 5 - 33}{2 \cdot 1} : \frac{- 5 - 33}{2}

The solutions to the quadratic equation are:

t = \frac{- 5 + 33}{2}, t = \frac{- 5 - 33}{2}

a^{2} + 7 a + 6

X - 5 = - 4 X + 15

\frac{x}{4} + \frac{x + 1}{3} = 0

- 3 x^{2} + 19 = 0

x^{2} - 2 x = - 5

What are the solutions to the equation t^2+5t-2=0 ?
The solutions to the equation t^2+5t-2=0 are t=(-5+sqrt(33))/2 ,t=(-5-sqrt(33))/2
Find the zeros of t^2+5t-2=0
The zeros of t^2+5t-2=0 are t=(-5+sqrt(33))/2 ,t=(-5-sqrt(33))/2