(3+4i)(2+7i)=

Solution for (3+4i)(2+7i)= equation:

(3+4i)(2+7i)=

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

(4i+3)(7i+2)=0

We multiply parentheses ..

(+28i^2+8i+21i+6)=0

We get rid of parentheses

28i^2+8i+21i+6=0

We add all the numbers together, and all the variables

28i^2+29i+6=0

a = 28; b = 29; c = +6;
Δ = b2-4ac
Δ = 292-4·28·6
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-13}{2*28}=\frac{-42}{56}
=-3/4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+13}{2*28}=\frac{-16}{56}
=-2/7
$