How do you solve the quadratic using the quadratic formula given $7-8z^2=6z+16$ over the set of complex numbers?

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Answer 1 · 2016-11-20T09:08:18

The solution is $S={(-3+3isqrt7)/(8),(-3-3isqrt7)/(8)}$

Rearranging the equation

$8z^2+6z+9=0$

Let's calculate the discriminant

$Delta=b^2-4ac=36-4*8*9=36-288=-252$

$Delta<0$ , so solutions are in $CC$

Then,
$z=(-b+-sqrtDelta)/(2a)=(-6+-isqrt252)/16$

$=(-6+-6isqrt7)/16=(-3+-3isqrt7)/(8)$

How do you solve the quadratic using the quadratic formula given 7-8z^2=6z+167-8z^2=6z+16 over the set of complex numbers?