Answer:
The quadratic equation has two complex solutions
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2}=4x-5[/tex]
Equate to cero
[tex]x^{2}-4x+5=0[/tex]
so
[tex]a=1\\b=-4\\c=5[/tex]
substitute in the formula
[tex]x=\frac{4(+/-)\sqrt{-4^{2}-4(1)(5)}} {2(1)}[/tex]
[tex]x=\frac{4(+/-)\sqrt{16-20}} {2}[/tex]
[tex]x=\frac{4(+/-)\sqrt{-4}} {2}[/tex]
Remember that
[tex]i^{2}=-1[/tex]
so
[tex]x=\frac{4(+/-)2i} {2}[/tex]
[tex]x=2(+/-)i[/tex]
therefore
The quadratic equation has two complex solutions
