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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-21i^2-14i+18i+12)=0

We get rid of parentheses

-21i^2-14i+18i+12=0

We add all the numbers together, and all the variables

-21i^2+4i+12=0

a = -21; b = 4; c = +12;
Δ = b2-4ac
Δ = 42-4·(-21)·12
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-32}{2*-21}=\frac{-36}{-42}
=6/7
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+32}{2*-21}=\frac{28}{-42}
=-2/3
$