Answer:
The roots are
[tex]x=-1+i\frac{\sqrt{10}}{2}[/tex]
[tex]x=-1-i\frac{\sqrt{10}}{2}[/tex]
Step-by-step explanation:
we have
[tex]2x^2+4x+7[/tex]
To find the roots equate the equation to zero
so
[tex]2x^2+4x+7=0[/tex]
we know that
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+4x+7=0[/tex]
so
[tex]a=2\\b=4\\c=7[/tex]
substitute in the formula
[tex]x=\frac{-4\pm\sqrt{4^{2}-4(2)(7)}} {2(2)}[/tex]
[tex]x=\frac{-4\pm\sqrt{-40}} {4}[/tex]
Remember that
[tex]i=\sqrt{-1}[/tex]
so
[tex]x=\frac{-4\pmi\sqrt{40}} {4}[/tex]
[tex]x=\frac{-4\pm2i\sqrt{10}} {4}[/tex]
simplify
[tex]x=\frac{-2\pm i\sqrt{10}} {2}[/tex]
therefore
The roots are
[tex]x=\frac{-2+i\sqrt{10}} {2}=-1+i\frac{\sqrt{10}}{2}[/tex]
[tex]x=\frac{-2-i\sqrt{10}} {2}=-1-i\frac{\sqrt{10}}{2}[/tex]
