Answer:
![x_{1} \approx -0.70](https://tex.z-dn.net/?f=x_%7B1%7D%20%5Capprox%20-0.70)
![x_{2} \approx -4.30](https://tex.z-dn.net/?f=x_%7B2%7D%20%5Capprox%20-4.30)
Step-by-step explanation:
The given equation is: ![x^{2}+5x+3=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B5x%2B3%3D0)
To find the solution of any quadratic equation we use:
![x_{1,2}=\frac{-b \±\sqrt{b^{2}-4ac} }{2a}](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%3D%5Cfrac%7B-b%20%5C%C2%B1%5Csqrt%7Bb%5E%7B2%7D-4ac%7D%20%7D%7B2a%7D)
Where:
is the coefficient of the quadratic term.
is the coefficient of the linear term.
is the independent term.
So, according to this, each variables is equal to:
![b=5](https://tex.z-dn.net/?f=b%3D5)
Now, we substitute these values in the formula:
![x_{1,2}=\frac{-5 \±\sqrt{(5)^{2}-4(1)(3)} }{2(1)}](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%3D%5Cfrac%7B-5%20%5C%C2%B1%5Csqrt%7B%285%29%5E%7B2%7D-4%281%29%283%29%7D%20%7D%7B2%281%29%7D)
![x_{1,2}=\frac{-5 \±\sqrt{25-12} }{2}](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%3D%5Cfrac%7B-5%20%5C%C2%B1%5Csqrt%7B25-12%7D%20%7D%7B2%7D)
![x_{1,2}=\frac{-5 \±\sqrt{13} }{2}](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%3D%5Cfrac%7B-5%20%5C%C2%B1%5Csqrt%7B13%7D%20%7D%7B2%7D)
![x_{1,2}=\frac{-5 \±\sqrt{13} }{2}](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%3D%5Cfrac%7B-5%20%5C%C2%B1%5Csqrt%7B13%7D%20%7D%7B2%7D)
So, one solution has the positive sign, and the other the negative sign. Therefore the solutions are:
![x_{1}=\frac{-5 +\sqrt{13} }{2} \approx -0.70](https://tex.z-dn.net/?f=x_%7B1%7D%3D%5Cfrac%7B-5%20%2B%5Csqrt%7B13%7D%20%7D%7B2%7D%20%5Capprox%20-0.70)
![x_{2}=\frac{-5 -\sqrt{13} }{2} \approx -4.30](https://tex.z-dn.net/?f=x_%7B2%7D%3D%5Cfrac%7B-5%20-%5Csqrt%7B13%7D%20%7D%7B2%7D%20%5Capprox%20-4.30)
As you can see, the solution was founded just by using the formula, identifying the values of a, b and c. Then, solving the formula we all values replaced.