Respuesta :
The easiest way it to solve it.
[tex] {x}^{2} + 5x + 2 = 0[/tex]
You have
[tex] \rho = 25 - 4 \times 1 \times 2 = 17[/tex]
Thus
[tex]x = \frac{ - 5 \pm \sqrt{17} }{2} [/tex]
[tex] {x}^{2} + 5x + 2 = 0[/tex]
You have
[tex] \rho = 25 - 4 \times 1 \times 2 = 17[/tex]
Thus
[tex]x = \frac{ - 5 \pm \sqrt{17} }{2} [/tex]
Answer:
[tex]x=\frac{-5+\sqrt{17}}{2}[/tex] and [tex]x=\frac{-5-\sqrt{17}}{2}[/tex]
Step-by-step explanation:
[tex]x^2 + 5x = -2[/tex]
To solve for x, make right hand side 0
Add 2 on both sides.
[tex]x^2 + 5x +2=0[/tex]
Apply quadratic formula to solve for x
a=1, b=5, c= 2
[tex]x=\frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]
Plug in the values and find out x
[tex]x=\frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-5+-\sqrt{5^2-4(1)(2)}}{2(1)}[/tex]
[tex]x=\frac{-5+-\sqrt{17}}{2}[/tex]
[tex]x=\frac{-5+\sqrt{17}}{2}[/tex] and [tex]x=\frac{-5-\sqrt{17}}{2}[/tex]
