What are the solutions to the equation 1x^2+4x+3=0 ?

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

Find the zeros of 1x^2+4x+3=0

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

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

x = - 1, x = - 3

Solution steps

1 \cdot x^{2} + 4 x + 3 = 0

Refine

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1, x = - 3

(x + 4) (x - 6) = 0

3 x^{2} = - 12 x

x, f = (x + y)^{2}

7 x^{2} - 31 x - 20 = 0

x^{2} - 0.5 x - 5 = 0

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