Solve for x over the complex numbers. x 2 +4x+13=0 Enter your answers in the boxes in standard form.

Answer:The solutions are x=-2+3i and x=-2-3iStep-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] thereforeThe solutions are x=-2+3i and x=-2-3i