Answer:
The solutions are x=-2+3i and x=-2-3i
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+13=0[/tex]
so
[tex]a=1\\b=4\\c=13[/tex]
substitute in the formula
[tex]x=\frac{-4(+/-)\sqrt{4^{2}-4(1)(13)}} {2(1)}[/tex]
[tex]x=\frac{-4(+/-)\sqrt{-36}} {2}[/tex]
Remember that
[tex]i^{2} =-1\\i=\sqrt{-1}[/tex]
substitute
[tex]x=\frac{-4(+/-)6i} {2}[/tex]
[tex]x_1=\frac{-4(+)6i} {2}=-2+3i[/tex]
[tex]x_2=\frac{-4(-)6i} {2}=-2-3i[/tex]
therefore
The solutions are x=-2+3i and x=-2-3i
