How do you solve the quadratic with complex numbers given $5x^2+12x+8=0$ ?

Question

2826 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-25T08:13:10

The solutions are $S={-6/5-2/5i,-6/5+2/5i}$

We need $ax^2+bx+c=0$

$Delta=b^2-4ac$

$x=(-b+-sqrtDelta)/(2a)$

The equation is $5x^2+12x+8=0$

$Delta=12^2-4*5*8=144-160=-16$

As, $Delta<0$ , there are no roots in $RR$ but in $CC$

$x=(-12+-4i)/10$

The roots are $x_1=-6/5-2/5i$ and $x_2=-6/5+2/5i$

How do you solve the quadratic with complex numbers given 5x^2+12x+8=05x^2+12x+8=0?