Answer:
x² + 1 = 0
a = 1
b = 0
c = 1
[tex]x1.2 = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a} [/tex]
[tex]x1.2 = \frac{ - 0 ±\sqrt{ {0}^{2} - (4.1.1) } }{2.1} [/tex]
[tex]x1.2 = \frac{0± \sqrt{0 - 4} }{2} [/tex]
[tex]x1.2 = \frac{± \sqrt{ - 4} }{2} [/tex]
There is no real solution because the discriminant is negative
Note.
Formula diskriminan
√b² - 4ac.
if the diskrimanan is negatif, so the quadratic equatoin dont have solution