What are the solutions to the equation t^2+4t+3=0 ?

The solutions to the equation t^2+4t+3=0 are t=-1,t=-3

Find the zeros of t^2+4t+3=0

The zeros of t^2+4t+3=0 are t=-1,t=-3

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t^{2} + 4 t + 3 = 0

t = - 1, t = - 3

Solution steps

t^{2} + 4 t + 3 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot 3 = 2

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

Separate the solutions

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

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

t = \frac{- 4 - 2}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

t = - 1, t = - 3

5 x^{2} - 85 = 40

x, x^{2} y - x^{2} + 4 y = 0

9 x^{2} + 5 x + 1 = 0

(x - 2)^{2} + (x - 4)^{2} = x^{2}

t^{2} + 10 t + 9 = 0

What are the solutions to the equation t^2+4t+3=0 ?
The solutions to the equation t^2+4t+3=0 are t=-1,t=-3
Find the zeros of t^2+4t+3=0
The zeros of t^2+4t+3=0 are t=-1,t=-3