1/2c+4=2c+5

Solution for 1/2c+4=2c+5 equation:

1/2c+4=2c+5

We move all terms to the left:

1/2c+4-(2c+5)=0


Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R


We get rid of parentheses

1/2c-2c-5+4=0

We multiply all the terms by the denominator


-2c*2c-5*2c+4*2c+1=0

Wy multiply elements

-4c^2-10c+8c+1=0

We add all the numbers together, and all the variables

-4c^2-2c+1=0

a = -4; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·(-4)·1
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{5}}{2*-4}=\frac{2-2\sqrt{5}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{5}}{2*-4}=\frac{2+2\sqrt{5}}{-8}
$