2c-8/c=10

Solution for 2c-8/c=10 equation:

2c-8/c=10

We move all terms to the left:

2c-8/c-(10)=0


Domain of the equation: c!=0

c∈R


We multiply all the terms by the denominator


2c*c-10*c-8=0

We add all the numbers together, and all the variables

-10c+2c*c-8=0

Wy multiply elements

2c^2-10c-8=0

a = 2; b = -10; c = -8;
Δ = b2-4ac
Δ = -102-4·2·(-8)
Δ = 164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{41}}{2*2}=\frac{10-2\sqrt{41}}{4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{41}}{2*2}=\frac{10+2\sqrt{41}}{4}
$