Answer:
[tex]x=-4+2\sqrt{5}[/tex]
[tex]x=-4-2\sqrt{5}[/tex]
Step-by-step explanation:
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]2x^{2} +16x-8=0[/tex]
so
[tex]a=2\\b=16\\c=-8[/tex]
substitute in the formula
[tex]x=\frac{-16(+/-)\sqrt{16^{2}-4(2)(-8)}} {2(2)}[/tex]
[tex]x=\frac{-16(+/-)\sqrt{320}} {4}[/tex]
[tex]x=\frac{-16(+/-)8\sqrt{5}} {4}[/tex]
[tex]x=-4(+/-)2\sqrt{5}[/tex]
therefore
The solutions are
[tex]x=-4+2\sqrt{5}[/tex]
[tex]x=-4-2\sqrt{5}[/tex]
