Respuesta :
Answer:
[tex]x_{1} \approx -0.70[/tex]
[tex]x_{2} \approx -4.30[/tex]
Step-by-step explanation:
The given equation is: [tex]x^{2}+5x+3=0[/tex]
To find the solution of any quadratic equation we use:
[tex]x_{1,2}=\frac{-b \±\sqrt{b^{2}-4ac} }{2a}[/tex]
Where:
[tex]a:[/tex] is the coefficient of the quadratic term.
[tex]b:[/tex] is the coefficient of the linear term.
[tex]c:[/tex] is the independent term.
So, according to this, each variables is equal to:
[tex]a=1[/tex]
[tex]b=5[/tex]
[tex]c=3[/tex]
Now, we substitute these values in the formula:
[tex]x_{1,2}=\frac{-5 \±\sqrt{(5)^{2}-4(1)(3)} }{2(1)}[/tex]
[tex]x_{1,2}=\frac{-5 \±\sqrt{25-12} }{2}[/tex]
[tex]x_{1,2}=\frac{-5 \±\sqrt{13} }{2}[/tex]
[tex]x_{1,2}=\frac{-5 \±\sqrt{13} }{2}[/tex]
So, one solution has the positive sign, and the other the negative sign. Therefore the solutions are:
[tex]x_{1}=\frac{-5 +\sqrt{13} }{2} \approx -0.70[/tex]
[tex]x_{2}=\frac{-5 -\sqrt{13} }{2} \approx -4.30[/tex]
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.
