we have
[tex]x^{2} -3x=-8[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]x^{2} -3x+1.5^{2}=-8+1.5^{2}[/tex]
[tex]x^{2} -3x+1.5^{2}=-5.75[/tex]
Rewrite as perfect squares
[tex](x-1.5)^{2}=-5.75[/tex]
Square Root both sides
[tex]x-1.5=(+/-)\sqrt{-5.75}[/tex]
Remember that
[tex]\sqrt{-1}=i[/tex]
[tex]\sqrt{5.75}= \sqrt{\frac{23}{4}}=\frac{\sqrt{23}}{2}[/tex]
Substitute
[tex]x-1.5=(+/-)i\frac{\sqrt{23}}{2}[/tex]
[tex]x=1.5(+/-)i\frac{\sqrt{23}}{2}[/tex]
[tex]x=(3/2)(+/-)i\frac{\sqrt{23}}{2}[/tex]
[tex]x=(3/2)+i\frac{\sqrt{23}}{2}=\frac{3+i\sqrt{23}}{2}[/tex]
[tex]x=(3/2)-i\frac{\sqrt{23}}{2}=\frac{3-i\sqrt{23}}{2}[/tex]
therefore
the answer is
x equals quantity of 3 plus or minus I square root of 23 all over 2
