What are the solutions to the equation t^2+4t+5=0 ?

The solutions to the equation t^2+4t+5=0 are t=-2+i,t=-2-i

Find the zeros of t^2+4t+5=0

The zeros of t^2+4t+5=0 are t=-2+i,t=-2-i

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t^{2} + 4 t + 5 = 0

t = - 2 + i, t = - 2 - i

Solution steps

t^{2} + 4 t + 5 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 5 : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = - 2 + i, t = - 2 - i

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

y, 2 x^{2} + 3 y^{2} = 0

x^{2} + 2 x + 82 = 0

9 x^{2} + 12 x + 5 = 1

x, x^{2} + x y = l n y

What are the solutions to the equation t^2+4t+5=0 ?
The solutions to the equation t^2+4t+5=0 are t=-2+i,t=-2-i
Find the zeros of t^2+4t+5=0
The zeros of t^2+4t+5=0 are t=-2+i,t=-2-i