0=(8+i)(2+7i)

Solution for 0=(8+i)(2+7i) equation:

0=(8+i)(2+7i)

We move all terms to the left:

0-((8+i)(2+7i))=0

We add all the numbers together, and all the variables

-((i+8)(7i+2))+0=0

We add all the numbers together, and all the variables

-((i+8)(7i+2))=0

We multiply parentheses ..

-((+7i^2+2i+56i+16))=0


We calculate terms in parentheses: -((+7i^2+2i+56i+16)), so:

(+7i^2+2i+56i+16)

We get rid of parentheses

7i^2+2i+56i+16

We add all the numbers together, and all the variables

7i^2+58i+16

Back to the equation:

-(7i^2+58i+16)


We get rid of parentheses

-7i^2-58i-16=0

a = -7; b = -58; c = -16;
Δ = b2-4ac
Δ = -582-4·(-7)·(-16)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-58)-54}{2*-7}=\frac{4}{-14}
=-2/7
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-58)+54}{2*-7}=\frac{112}{-14}
=-8
$