7x+(42/x)=63

Solution for 7x+(42/x)=63 equation:

7x+(42/x)=63

We move all terms to the left:

7x+(42/x)-(63)=0


Domain of the equation: x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

7x+(+42/x)-63=0

We get rid of parentheses

7x+42/x-63=0

We multiply all the terms by the denominator


7x*x-63*x+42=0

We add all the numbers together, and all the variables

-63x+7x*x+42=0

Wy multiply elements

7x^2-63x+42=0

a = 7; b = -63; c = +42;
Δ = b2-4ac
Δ = -632-4·7·42
Δ = 2793
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2793}=\sqrt{49*57}=\sqrt{49}*\sqrt{57}=7\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-63)-7\sqrt{57}}{2*7}=\frac{63-7\sqrt{57}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-63)+7\sqrt{57}}{2*7}=\frac{63+7\sqrt{57}}{14}
$