COMPLETING THE SQUARE

Answer:13. [tex]x=1,-7[/tex]14. [tex]x=-3+\sqrt{14}[/tex] and [tex]x=-3-\sqrt{14}[/tex]Step-by-step explanation:to solve the quadratic equation([tex]ax^{2} +bx+c=0[/tex]) using completing the squares method:Step1: divide the equation by a to make it in the form [tex]x^{2} +\frac{b}{a}x+\frac{c}{a} =0[/tex]Step2: add [tex](\frac{b}{2a})^{2}[/tex] on both sides of the equation to get the eqaution:[tex]x^{2} +\frac{b}{a}x+(\frac{b}{2a})^{2} +\frac{c}{a}=(\frac{b}{2a})^{2}[/tex]Step3: rearrange them to get the square.⇒[tex](x+\frac{b}{2a} )^{2}=(\frac{b}{2a})^{2}-\frac{c}{a}[/tex]⇒[tex]x= -\frac{b}{2a}+\sqrt{(\frac{b}{2a})^{2}-\frac{c}{a}}[/tex] and [tex]x= \frac{b}{2a}+\sqrt{(\frac{b}{2a})^{2}-\frac{c}{a}}[/tex]Now getting on to the question:13. [tex]x^{2} +6x=7[/tex]a=1; b=6; c=-7adding [tex]\frac{b}{2a}^{2} = \frac{6}{2*1}^{2} =3^{2} =9[/tex] on both sides⇒[tex]x^{2} +6x+9=7+9[/tex]⇒[tex]x^{2} +6x+9=16[/tex]⇒[tex](x+3)^{2}=16[/tex]⇒[tex]x+3=\sqrt{16}[/tex]⇒[tex]x+3=4[/tex] and [tex]x+3=-4[/tex]⇒[tex]x=1,-7[/tex]14. [tex]x^{2} +6x=5[/tex]a=1; b=6; c=-5adding  [tex]\frac{b}{2a}^{2} = \frac{6}{2*1}^{2} =3^{2} =9[/tex]on both sides⇒[tex]x^{2} +6x+9=5+9[/tex]⇒[tex]x^{2} +6x+9=14[/tex]⇒[tex](x+3)^{2}=14[/tex]⇒[tex]x+3=\sqrt{14}[/tex]⇒[tex]x+3=\sqrt{14}[/tex] and [tex]x+3=-\sqrt{14}[/tex]⇒[tex]x=-3+\sqrt{14}[/tex] and [tex]x=-3-\sqrt{14}[/tex]