Completing the Square- Color By Number Thanksgiving Edition

Answer: The roots of question 1) -6 , 2 (Brown) 2) -7, 1  (light brown)3) 3, -1  (yellow)4) [tex]\frac{2\pm \sqrt{19} }{3}[/tex] (orange)5)[tex]3\pm \sqrt{15}[/tex] (Red)6)[tex]\frac{-5\pm3\sqrt{5} }{2}[/tex]  (Brown)Step-by-step explanation:Since, First quadratic equation is, [tex]x^2+4x-12=0[/tex]⇒[tex]x^2+6x-2x-12=0[/tex]⇒[tex]x(x+6)-2(x+6)=0[/tex]⇒[tex](x-2)(x+6)=0[/tex]Therefore, root of  [tex]x^2+4x-12=0[/tex] are x = 2, -6 Now,Second quadratic equation is, [tex]2x^2+12x-14=0[/tex]⇒[tex]2x^2+14x-2x-14=0[/tex]⇒[tex]2x(x+7)-2(x+7)=0[/tex]⇒[tex](2x-2)(x+7)=0[/tex]Therefore, root of  [tex]2x^2+12x-14=0[/tex] are x = 1, -7 Third quadratic equation is, [tex]x^2-2x-3=0[/tex]⇒[tex]x^2-3x+x-3=0[/tex]⇒[tex]x(x-3)+1(x-3)=0[/tex]⇒[tex](x-3)(x+1)=0[/tex]Therefore, root of  [tex]2x^2+12x-14=0[/tex] are x = -1, 3fourth quadratic equation is,  [tex]3x^2-4x-5=0[/tex]By applying quadratic formula, roots of the equation [tex]3x^2-4x-5=0[/tex]are[tex]x=\frac{2\pm 19}{3}[/tex]     ([tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a }[/tex])Similarly, By the quadratic formula,The roots of [tex]x^2-6x+4=0[/tex] are [tex]3\pm \sqrt{15}[/tex]And, roots of [tex]3x^2-4x-5=0[/tex] are [tex]\frac{-5\pm3\sqrt{5} }{2}[/tex]