-i(6+7i)-3i(2-5i)=1

Solution for -i(6+7i)-3i(2-5i)=1 equation:

-i(6+7i)-3i(2-5i)=1

We move all terms to the left:

-i(6+7i)-3i(2-5i)-(1)=0

We add all the numbers together, and all the variables

-i(7i+6)-3i(-5i+2)-1=0

We multiply parentheses

-7i^2+15i^2-6i-6i-1=0

We add all the numbers together, and all the variables

8i^2-12i-1=0

a = 8; b = -12; c = -1;
Δ = b2-4ac
Δ = -122-4·8·(-1)
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{11}}{2*8}=\frac{12-4\sqrt{11}}{16}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{11}}{2*8}=\frac{12+4\sqrt{11}}{16}
$