Answer:
The zeros are
[tex]x=-2+\frac{\sqrt{22}}{2}[/tex]
[tex]x=-2-\frac{\sqrt{22}}{2}[/tex]
Step-by-step explanation:
we have
[tex]f(x)=2x^{2}+8x-3[/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
Equate the function to zero
[tex]2x^{2} +8x-3=0[/tex]
so
[tex]a=2\\b=8\\c=-3[/tex]
substitute in the formula
[tex]x=\frac{-8\pm\sqrt{8^{2}-4(2)(-3)}} {2(2)}[/tex]
[tex]x=\frac{-8\pm\sqrt{88}} {4}[/tex]
[tex]x=\frac{-8\pm2\sqrt{22}} {4}[/tex]
[tex]x=-2\pm\frac{\sqrt{22}}{2}[/tex]
therefore
[tex]x=-2+\frac{\sqrt{22}}{2}[/tex]
[tex]x=-2-\frac{\sqrt{22}}{2}[/tex]
