-62i+21i^2=-16

Solution for -62i+21i^2=-16 equation:

-62i+21i^2=-16

We move all terms to the left:

-62i+21i^2-(-16)=0

We add all the numbers together, and all the variables

21i^2-62i+16=0

a = 21; b = -62; c = +16;
Δ = b2-4ac
Δ = -622-4·21·16
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-62)-50}{2*21}=\frac{12}{42}
=2/7
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+50}{2*21}=\frac{112}{42}
=2+2/3
$