(7k-3)=(k2-2k+7)

Solution for (7k-3)=(k2-2k+7) equation:

(7k-3)=(k2-2k+7)

We move all terms to the left:

(7k-3)-((k2-2k+7))=0

We add all the numbers together, and all the variables

-((+k^2-2k+7))+(7k-3)=0

We get rid of parentheses

-((+k^2-2k+7))+7k-3=0


We calculate terms in parentheses: -((+k^2-2k+7)), so:

(+k^2-2k+7)

We get rid of parentheses

k^2-2k+7

Back to the equation:

-(k^2-2k+7)


We add all the numbers together, and all the variables

7k-(k^2-2k+7)-3=0

We get rid of parentheses

-k^2+7k+2k-7-3=0

We add all the numbers together, and all the variables

-1k^2+9k-10=0

a = -1; b = 9; c = -10;
Δ = b2-4ac
Δ = 92-4·(-1)·(-10)
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{41}}{2*-1}=\frac{-9-\sqrt{41}}{-2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{41}}{2*-1}=\frac{-9+\sqrt{41}}{-2}
$