3i(2+5i)+(6-7i)-(9+i)=0

Solution for 3i(2+5i)+(6-7i)-(9+i)=0 equation:

3i(2+5i)+(6-7i)-(9+i)=0

We add all the numbers together, and all the variables

3i(5i+2)+(-7i+6)-(i+9)=0

We multiply parentheses

15i^2+6i+(-7i+6)-(i+9)=0

We get rid of parentheses

15i^2+6i-7i-i+6-9=0

We add all the numbers together, and all the variables

15i^2-2i-3=0

a = 15; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·15·(-3)
Δ = 184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{46}}{2*15}=\frac{2-2\sqrt{46}}{30}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{46}}{2*15}=\frac{2+2\sqrt{46}}{30}
$