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

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

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

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

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

y = - 1, y = - 3

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = - 1, y = - 3

6 x^{2} + 14 = 19 x

\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}

6.25 x + 7.5 - 2.5 x = 3.75 x - 8.75

2 + \frac{1}{4}

∣ x - 177.2612 ∣ \leq 4.57

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