(2-i)(3+4i)=0

Solution for (2-i)(3+4i)=0 equation:

(2-i)(3+4i)=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4i^2-3i+8i+6)=0

We get rid of parentheses

-4i^2-3i+8i+6=0

We add all the numbers together, and all the variables

-4i^2+5i+6=0

a = -4; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·(-4)·6
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*-4}=\frac{-16}{-8}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*-4}=\frac{6}{-8}
=-3/4
$