What are the solutions to the equation y^2+4y+13=0 ?

The solutions to the equation y^2+4y+13=0 are y=-2+3i,y=-2-3i

Find the zeros of y^2+4y+13=0

The zeros of y^2+4y+13=0 are y=-2+3i,y=-2-3i

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y^{2} + 4 y + 13 = 0

y = - 2 + 3 i, y = - 2 - 3 i

Solution steps

y^{2} + 4 y + 13 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 13 : 6 i

y_{1, 2} = \frac{- 4 \pm 6 i}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 4 + 6 i}{2 \cdot 1}, y_{2} = \frac{- 4 - 6 i}{2 \cdot 1}

y = \frac{- 4 + 6 i}{2 \cdot 1} : - 2 + 3 i

y = \frac{- 4 - 6 i}{2 \cdot 1} : - 2 - 3 i

The solutions to the quadratic equation are:

y = - 2 + 3 i, y = - 2 - 3 i

∣ x + 5 ∣ = ∣ 8 x + 3 ∣

(x - 5) + (3 x + 7) = x - (2 - 3 x)

1 - e^{- 0.64 \cdot \frac{89828}{2 \cdot 0.815}}

3 x^{2} (x^{2} - 16) \leq 0

15 \cdot \frac{3}{5}

What are the solutions to the equation y^2+4y+13=0 ?
The solutions to the equation y^2+4y+13=0 are y=-2+3i,y=-2-3i
Find the zeros of y^2+4y+13=0
The zeros of y^2+4y+13=0 are y=-2+3i,y=-2-3i