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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We add all the numbers together, and all the variables

8i^2-12i=0

a = 8; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·8·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*8}=\frac{0}{16}
=0
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*8}=\frac{24}{16}
=1+1/2
$