What are the solutions to the equation t^2+4t-2=0 ?

The solutions to the equation t^2+4t-2=0 are t=-2+sqrt(6),t=-2-sqrt(6)

Find the zeros of t^2+4t-2=0

The zeros of t^2+4t-2=0 are t=-2+sqrt(6),t=-2-sqrt(6)

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t^{2} + 4 t - 2 = 0

t = - 2 + 6, t = - 2 - 6

Decimal Notation

t = 0.44948 \dots, t = - 4.44948 \dots

Solution steps

t^{2} + 4 t - 2 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 2) = 26

t_{1, 2} = \frac{- 4 \pm 2 6}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 4 + 2 6}{2 \cdot 1}, t_{2} = \frac{- 4 - 2 6}{2 \cdot 1}

t = \frac{- 4 + 2 6}{2 \cdot 1} : - 2 + 6

t = \frac{- 4 - 2 6}{2 \cdot 1} : - 2 - 6

The solutions to the quadratic equation are:

t = - 2 + 6, t = - 2 - 6

b = b^{2} - 2

y^{2} + 9 y + 14 = 0

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

x, k x^{2} - 17 x + 33 = 0

1 2^{2} = m \cdot m + 7

What are the solutions to the equation t^2+4t-2=0 ?
The solutions to the equation t^2+4t-2=0 are t=-2+sqrt(6),t=-2-sqrt(6)
Find the zeros of t^2+4t-2=0
The zeros of t^2+4t-2=0 are t=-2+sqrt(6),t=-2-sqrt(6)