Answer
$z=i\sqrt 5/2$, $z=i\sqrt 5/-2$, $z=i\sqrt 2/2$, $z=i\sqrt 2/-2$
Work Step by Step
$8z^4+14z^2=-5$
Let $x=z^2$
$8z^4+14z^2=-5$
$8(z^2*z^2)+14(z^2)=-5$
$8x^2+14x=-5$
$8x^2+14x+5=-5+5$
$8x^2+14x+5=0$
Let $a=8$, $b=14$, $c=5$
$x=(−b±\sqrt{b^2−4ac})/2a$
$x=(−14±\sqrt{14^2−4*8*5})/2∗8$
$x=(−14±\sqrt {196-160})/16$
$x=(−14±\sqrt{36}/16$
$x=(−14±6)/16$
$x=(-14-6)/16$
$x=-20/16$
$x=-5/4$
$x=(-14+6)/16$
$x=-8/16$
$x=-1/2$
$z^2=-5/4, -1/2$
$z^2=-5/4$
$\sqrt {z^2} = \sqrt {-5/4}$
$z = \sqrt {-5} /\sqrt 4$
$z = \sqrt {-1*5} /±2$
$z=i\sqrt 5/±2$
$z^2=-1/2$
$\sqrt {z^2} = \sqrt {-1/2}$
$z = \sqrt {-1}*\sqrt 2 /\sqrt 2*\sqrt 2$
$z = i\sqrt {1*2} /±2$
$z=i\sqrt 2/±2$
