Answer:
[tex]\boxed{\sf \ \ \ x=-4.12 \ or \ x=0.12 \ \ \ }[/tex]
Step-by-step explanation:
Hello,
step 1 - we divide all terms by 2
[tex]2x^2+8x-1=0 <=> x^2+4x-\dfrac{1}{2}=0[/tex]
step 2 - we complete the square
we can notice that
[tex]x^2+4x=(x+2)^2-4[/tex]
so
[tex]2x^2+8x-1=0 <=> x^2+4x-\dfrac{1}{2}=0<=>(x+2)^2-4-\dfrac{1}{2}=0\\[/tex]
step 3 - we move the constant term to the right of the equation
[tex](x+2)^2-4-\dfrac{1}{2}=0\\\\<=> (x+2)^2=4+\dfrac{1}{2}=\dfrac{8+1}{2}=\dfrac{9}{2}[/tex]
step 4 - we take the square root on both sides of the equation
[tex]x+2=\sqrt{\dfrac{9}{2}}[/tex]
or
[tex]x+2=-\sqrt{\dfrac{9}{2}}[/tex]
step 5 - we subtract 2 from both sides
[tex]x+2=\sqrt{\dfrac{9}{2}}<=> x=\dfrac{3}{\sqrt{2}}-2=0.12132...[/tex]
or
[tex]x+2=-\sqrt{\dfrac{9}{2}}<=> x=-\dfrac{3}{\sqrt{2}}-2=-4.12132...[/tex]
so the solutions are 0.12 and -4.12
hope this helps
