I^2-(2-i)i=(1-7i)

Solution for I^2-(2-i)i=(1-7i) equation:

^2-(2-I)I=(1-7I)

We move all terms to the left:

^2-(2-I)I-((1-7I))=0

We add all the numbers together, and all the variables

-(-1I+2)I-((-7I+1))+^2=0

We add all the numbers together, and all the variables

-(-1I+2)I-((-7I+1))=0

We multiply parentheses

1I^2-2I-((-7I+1))=0


We calculate terms in parentheses: -((-7I+1)), so:

(-7I+1)

We get rid of parentheses

-7I+1

Back to the equation:

-(-7I+1)


We add all the numbers together, and all the variables

I^2-2I-(-7I+1)=0

We get rid of parentheses

I^2-2I+7I-1=0

We add all the numbers together, and all the variables

I^2+5I-1=0

a = 1; b = 5; c = -1;
Δ = b2-4ac
Δ = 52-4·1·(-1)
Δ = 29
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{29}}{2*1}=\frac{-5-\sqrt{29}}{2}
$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{29}}{2*1}=\frac{-5+\sqrt{29}}{2}
$