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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+6i^2+15i-14i-35)=0

We get rid of parentheses

6i^2+15i-14i-35=0

We add all the numbers together, and all the variables

6i^2+i-35=0

a = 6; b = 1; c = -35;
Δ = b2-4ac
Δ = 12-4·6·(-35)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-29}{2*6}=\frac{-30}{12}
=-2+1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+29}{2*6}=\frac{28}{12}
=2+1/3
$