1-7k-6=3/7k+4

Solution for 1-7k-6=3/7k+4 equation:

1-7k-6=3/7k+4

We move all terms to the left:

1-7k-6-(3/7k+4)=0


Domain of the equation: 7k+4)!=0

k∈R


We add all the numbers together, and all the variables

-7k-(3/7k+4)-5=0

We get rid of parentheses

-7k-3/7k-4-5=0

We multiply all the terms by the denominator


-7k*7k-4*7k-5*7k-3=0

Wy multiply elements

-49k^2-28k-35k-3=0

We add all the numbers together, and all the variables

-49k^2-63k-3=0

a = -49; b = -63; c = -3;
Δ = b2-4ac
Δ = -632-4·(-49)·(-3)
Δ = 3381
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3381}=\sqrt{49*69}=\sqrt{49}*\sqrt{69}=7\sqrt{69}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-63)-7\sqrt{69}}{2*-49}=\frac{63-7\sqrt{69}}{-98}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-63)+7\sqrt{69}}{2*-49}=\frac{63+7\sqrt{69}}{-98}
$