What are the solutions to the equation t^2+8t+19=0 ?

The solutions to the equation t^2+8t+19=0 are t=-4+sqrt(3)i,t=-4-sqrt(3)i

Find the zeros of t^2+8t+19=0

The zeros of t^2+8t+19=0 are t=-4+sqrt(3)i,t=-4-sqrt(3)i

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t^{2} + 8 t + 19 = 0

t = - 4 + 3 i, t = - 4 - 3 i

Solution steps

t^{2} + 8 t + 19 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 19}{2 \cdot 1}

Simplify

8^{2} - 4 \cdot 1 \cdot 19 : 23 i

t_{1, 2} = \frac{- 8 \pm 2 3 i}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 8 + 2 3 i}{2 \cdot 1}, t_{2} = \frac{- 8 - 2 3 i}{2 \cdot 1}

t = \frac{- 8 + 2 3 i}{2 \cdot 1} : - 4 + 3 i

t = \frac{- 8 - 2 3 i}{2 \cdot 1} : - 4 - 3 i

The solutions to the quadratic equation are:

t = - 4 + 3 i, t = - 4 - 3 i

16 (x^{2} - 4) + 5 = - 10

7 x^{2} + 42 x - 50 = 6

0 = 32 - 2 t - 5 t^{2}

2 (x + 1)^{2} - 50 = 0

2 x^{2} + 4 = 5 x + 2

What are the solutions to the equation t^2+8t+19=0 ?
The solutions to the equation t^2+8t+19=0 are t=-4+sqrt(3)i,t=-4-sqrt(3)i
Find the zeros of t^2+8t+19=0
The zeros of t^2+8t+19=0 are t=-4+sqrt(3)i,t=-4-sqrt(3)i