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

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

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

We multiply parentheses ..

(+6i^2-21i+4i-14)=0

We get rid of parentheses

6i^2-21i+4i-14=0

We add all the numbers together, and all the variables

6i^2-17i-14=0

a = 6; b = -17; c = -14;
Δ = b2-4ac
Δ = -172-4·6·(-14)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-25}{2*6}=\frac{-8}{12}
=-2/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+25}{2*6}=\frac{42}{12}
=3+1/2
$