Question

For what values of $a$, that the roots of the equation $x^2+2(3a+5)x+2(9a^2+25)=0$ are complex

Answer 1

Roots of the quadratic equation $A{x}^2+Bx+C=0$ are complex if

$B^2-4AC<0$

For the given equation $x^2+2(3a+5)x+2(9a^2+25)=0$

$B^2-4AC=4(3a+5{)}^2-4(2\left)\right(9a^2+25)$

$=4(9a^2+25+30a)-8(9a^2+25)$

$=-4(9a^2+25-30a)$

$=-4(3a-5{)}^2$

which will be zero if $a=\frac{5}{3}$

or will be negative for any other value of $a$

So the roots are real and equal if $a=\frac{5}{3}$

and the roots are complex for any other value of $a$