Answer:
option A and B
[tex]x=\frac{-7+\sqrt{17}} {4}[/tex]
and
[tex]x=\frac{-7-\sqrt{17}} {4}[/tex]
Step-by-step explanation:
we have
[tex]2x^2+7x+4=0[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]2x^2+7x+4=0[/tex]
so
[tex]a=2\\b=7\\c=4[/tex]
substitute in the formula
[tex]x=\frac{-7\pm\sqrt{7^{2}-4(2)(4)}} {2(2)}[/tex]
[tex]x=\frac{-7\pm\sqrt{17}} {4}[/tex]
so
[tex]x=\frac{-7+\sqrt{17}} {4}[/tex]
and
[tex]x=\frac{-7-\sqrt{17}} {4}[/tex]
