(3-i)(4+7i)+3i(1-i)-1=0

Solution for (3-i)(4+7i)+3i(1-i)-1=0 equation:

(3-i)(4+7i)+3i(1-i)-1=0

We add all the numbers together, and all the variables

(-1i+3)(7i+4)+3i(-1i+1)-1=0

We multiply parentheses

-3i^2+(-1i+3)(7i+4)+3i-1=0

We multiply parentheses ..

-3i^2+(-7i^2-4i+21i+12)+3i-1=0

We get rid of parentheses

-3i^2-7i^2-4i+21i+3i+12-1=0

We add all the numbers together, and all the variables

-10i^2+20i+11=0

a = -10; b = 20; c = +11;
Δ = b2-4ac
Δ = 202-4·(-10)·11
Δ = 840
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:

$\sqrt{\Delta}=\sqrt{840}=\sqrt{4*210}=\sqrt{4}*\sqrt{210}=2\sqrt{210}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{210}}{2*-10}=\frac{-20-2\sqrt{210}}{-20}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{210}}{2*-10}=\frac{-20+2\sqrt{210}}{-20}
$