4i^2=-28i

Solution for 4i^2=-28i equation:

4i^2=-28i

We move all terms to the left:

4i^2-(-28i)=0

We get rid of parentheses

4i^2+28i=0

a = 4; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·4·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*4}=\frac{-56}{8}
=-7
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*4}=\frac{0}{8}
=0
$