Rewrite the right side of the equation
[tex]2x^2-4x-5 = 0[/tex] in the following way:
[tex]2x^2-4x-5 = 2(x^2-2x)-5=2(x^2-2x+1-1)-5=2((x-1)^2-1)-5=2(x-1)^2-2-5=2(x-2)^2-7[/tex].
Then the equation is [tex]2(x-2)^2-7=0[/tex] and [tex]2(x-2)^2=7[/tex].
So, [tex](x-2)^2= \frac{7}{2} [/tex] and [tex]x-2=\pm \sqrt{ \frac{7}{2} }[/tex]
The equation has two solutions:[tex]x_1=2+ \sqrt{ \frac{7}{2} } [/tex] and [tex]x_2=2- \sqrt{ \frac{7}{2} } [/tex].
