we have
[tex]f(x)=2x^{2}-10x-3[/tex]
we know that
To find the zeros of the quadratic equation equate to zero
[tex]2x^{2}-10x-3=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]2x^{2}-10x=3[/tex]
Factor the leading coefficient
[tex]2(x^{2}-5x)=3[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side.
[tex]2(x^{2}-5x+6.25)=3+12.5[/tex]
[tex]2(x^{2}-5x+6.25)=15.5[/tex]
Rewrite as perfect squares
[tex]2(x-2.5)^{2}=15.5[/tex]
[tex](x-2.5)^{2}=7.75[/tex]
[tex](x-2.5)=(+/-)\sqrt{7.75}[/tex]
[tex](x-2.5)=(+/-)2.78[/tex]
[tex]x1=(+)2.78+2.5=5.284[/tex] or [tex]x1=\frac{\sqrt{31}}{2} +2.5[/tex]
[tex]x2=(-)2.78+2.5=-0.284[/tex] or [tex]x2=-\frac{\sqrt{31}}{2} +2.5[/tex]
therefore
the answer is
[tex]x1=5.284[/tex] or [tex]x1=\frac{\sqrt{31}}{2} +2.5[/tex]
[tex]x2=-0.284[/tex] or [tex]x2=-\frac{\sqrt{31}}{2} +2.5[/tex]
see the attached figure to verify
