What are the solutions to the equation t^2+4t+1=3 ?

The solutions to the equation t^2+4t+1=3 are t=-2+sqrt(6),t=-2-sqrt(6)

Find the zeros of t^2+4t+1=3

The zeros of t^2+4t+1=3 are t=-2+sqrt(6),t=-2-sqrt(6)

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t^{2} + 4 t + 1 = 3

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

Decimal Notation

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

Solution steps

t^{2} + 4 t + 1 = 3

Move

3

to the left side

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

5 a^{2} + 7 = - 60

x^{2} - 4 x - 5 = 0

9 x^{2} - 5 = 607

x^{2} + 4 x = 0

- 4 x^{2} - 3 x + 2 = 0

What are the solutions to the equation t^2+4t+1=3 ?
The solutions to the equation t^2+4t+1=3 are t=-2+sqrt(6),t=-2-sqrt(6)
Find the zeros of t^2+4t+1=3
The zeros of t^2+4t+1=3 are t=-2+sqrt(6),t=-2-sqrt(6)