find the discriminant, and determine the number of real solutions. then solve x^2 +8x+20=0

Answer:Part 1) The quadratic equation has zero real solutionsPart 2) The solutions are[tex]x_1=-4+2i[/tex]   and [tex]x_2=-4-2i[/tex] Step-by-step explanation:we know thatThe 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}+8x+20=0[/tex]  so [tex]a=1\\b=8\\c=20[/tex] The discriminant is equal to[tex]D=(b^{2}-4ac)[/tex]If D=0 -----> the quadratic equation has only one real solutionIf D>0 -----> the quadratic equation has two real solutionsIf D<0 -----> the quadratic equation has two complex solutionsFind the value of D[tex]D=8^{2}-4(1)(20)=-16[/tex] -----> the quadratic equation has two complex solutionsFind out the solutionssubstitute the values of a,b and c in the formula[tex]x=\frac{-8(+/-)\sqrt{8^{2}-4(1)(20)}} {2(1)}[/tex] [tex]x=\frac{-8(+/-)\sqrt{-16}} {2}[/tex] Remember that[tex]i=\sqrt{-1}[/tex][tex]x=\frac{-8(+/-)4i} {2}[/tex] [tex]x_1=\frac{-8(+)4i} {2}=-4+2i[/tex] [tex]x_2=\frac{-8(-)4i} {2}=-4-2i[/tex]