Which of the following polynomials has solutions that are not real numbers?A) x^2-6x+3B) x^2+4x+3C)-x^2-9x-10D) x^2+2x+3

Answer:D) [tex]x^{2} + 2x + 3[/tex].Step-by-step explanation:All four polynomials are quadratic, meaning that the highest power of the unknown [tex]x[/tex] in the equation is two.The sign of the quadratic discriminant, [tex]\Delta[/tex], is a way to tell if a quadratic polynomial comes with non-real solutions.There are three cases:[tex]\Delta > 0[/tex]. The quadratic discriminant is positive. There are two real solutions and no non-real solution. The two solutions are different from each other.[tex]\Delta = 0[/tex]. The quadratic discriminant is zero. There is one real solution and no non-real solution.[tex]\Delta <0[/tex]. The quadratic discriminant is negative. There is no real solution and two non-real solutions.How to find the quadratic discriminant?If the equation is in this form:[tex]a \; x^{2} + b\;x + c = 0[/tex],where a, b, and c are real numbers (a.k.a. "constants.")Quadratic discriminant:[tex]\Delta = {b^{2} - 4\;a\cdot c}[/tex].Polynomial in A:[tex]x^{2} - 6x + 3 = 0[/tex].[tex]a = 1[/tex].[tex]b = -6[/tex].[tex]c = 3[/tex].[tex]\Delta = b^{2} - 4 \;a\cdot c = (-6)^{2} - 4\times 1\times 3 = 36 - 12 = 24[/tex].[tex]\Delta > 0[/tex]. There will be no non-real solutions and two distinct real solutions.Try the steps above for the polynomial in B, C, and D.B): [tex]\Delta = 4[/tex]. Two distinct real solutions. No non-real solution.C): [tex]\Delta= 41[/tex]. Two distinct real solutions. No non-real solution.D): [tex]\Delta = -8[/tex]. No real solution. Two distinct non-real solutions.