What are the solutions to the equation t^2+6t+4=0 ?

The solutions to the equation t^2+6t+4=0 are t=-3+sqrt(5),t=-3-sqrt(5)

Find the zeros of t^2+6t+4=0

The zeros of t^2+6t+4=0 are t=-3+sqrt(5),t=-3-sqrt(5)

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t^{2} + 6 t + 4 = 0

t = - 3 + 5, t = - 3 - 5

Decimal Notation

t = - 0.76393 \dots, t = - 5.23606 \dots

Solution steps

t^{2} + 6 t + 4 = 0

Solve with the quadratic formula

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

6^{2} - 4 \cdot 1 \cdot 4 = 25

t_{1, 2} = \frac{- 6 \pm 2 5}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 6 + 2 5}{2 \cdot 1}, t_{2} = \frac{- 6 - 2 5}{2 \cdot 1}

t = \frac{- 6 + 2 5}{2 \cdot 1} : - 3 + 5

t = \frac{- 6 - 2 5}{2 \cdot 1} : - 3 - 5

The solutions to the quadratic equation are:

t = - 3 + 5, t = - 3 - 5

∣ 5 x + 3 ∣ \geq - 2

- 9 < \frac{3 w}{8}

- 3 = (- x - 5) + 3

3 x - 5 > \frac{1}{2} x + 8 x

What are the solutions to the equation t^2+6t+4=0 ?
The solutions to the equation t^2+6t+4=0 are t=-3+sqrt(5),t=-3-sqrt(5)
Find the zeros of t^2+6t+4=0
The zeros of t^2+6t+4=0 are t=-3+sqrt(5),t=-3-sqrt(5)