$$x^2-2(3m-1)x+2m+3=0$$ Find the sum of solutions. It says that the sum equals to $-1$. I just can't wrap my head around this? Any help? Thx
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$\begingroup$In a quadratic equation $ax^2+bx+c=0$ the sum of the solutions (if any) is equal to $-b/a$. In fact, if $x_1$ and $x_2$ are the solutions then $$ax^2+bx+c=a(x-x_1)(x-x_2)=ax^2-a(x_1+x_2)x+a(x_1\cdot x_2)$$ and by comparing the coefficients we get $x_1+x_2=-b/a$.
In your case $x_1+x_2=2(3m-1)$. It seems that the sum of the (complex) solutions is $-1$ if $m=1/6$.
$\endgroup$ $\begingroup$$$x_{1,2}=3m-1\pm\sqrt{9m^2-8m-2}$$ therefore $$x_1+x_2=2(3m-1)$$
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