Question

How many real number solutions does 4x^2+2x+5=0 have?

Answer 1

Therefore $4x^2+2x+5=0$ have two real number solution.

Step-by-step explanation:

$4x^2+2x+5=0$                      The solution of the equation $ax^2+bx+c=0$ is

$\Leftrightarrow x = \frac{-2\pm \sqrt{2^2+4.4.5} }{2.5}$                     $x= \frac{-b\pm\sqrt{b^2+4ac} }{2a}$

$\Leftrightarrow x =\frac{-2\pm \sqrt{84} }{10}$

$\Leftrightarrow x =\frac{-1\pm \sqrt{21} }{5}$

Therefore $4x^2+2x+5=0$ have two real number solution