Answer:
C
Step-by-step explanation:
To solve an equation of the form [tex]ax^2+bx+c=0[/tex]
Use completing the square formula:
[tex]a\left(x+\dfrac{b}{2a} \right)^2+c-\dfrac{b^2}{4a}=0[/tex]
Given equation:
[tex]2x^2+16x-8=0[/tex]
Therefore,
Substitute these values into the formula:
[tex]2\left(x+\dfrac{16}{2(2)} \right)^2+(-8)-\dfrac{16^2}{4(2)}=0[/tex]
[tex]2\left(x+4} \right)^2-40=0[/tex]
To solve, add 40 to both sides:
[tex]2(x+4)^2=40[/tex]
Divide both sides by 2:
[tex](x+4)^2=20[/tex]
Square root both sides:
[tex]x+4=\pm \sqrt{20}[/tex]
[tex]x+4=\pm \sqrt{4 \cdot 5}[/tex]
[tex]x+4=\pm 2\sqrt{5}[/tex]
subtract 4 from both sides:
[tex]x=-4\pm 2\sqrt{5}[/tex]
