5(4c-1)=3c(2c-2)

Solution for 5(4c-1)=3c(2c-2) equation:

5(4c-1)=3c(2c-2)

We move all terms to the left:

5(4c-1)-(3c(2c-2))=0

We multiply parentheses

20c-(3c(2c-2))-5=0


We calculate terms in parentheses: -(3c(2c-2)), so:

3c(2c-2)

We multiply parentheses

6c^2-6c

Back to the equation:

-(6c^2-6c)


We get rid of parentheses

-6c^2+20c+6c-5=0

We add all the numbers together, and all the variables

-6c^2+26c-5=0

a = -6; b = 26; c = -5;
Δ = b2-4ac
Δ = 262-4·(-6)·(-5)
Δ = 556
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:

$\sqrt{\Delta}=\sqrt{556}=\sqrt{4*139}=\sqrt{4}*\sqrt{139}=2\sqrt{139}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{139}}{2*-6}=\frac{-26-2\sqrt{139}}{-12}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{139}}{2*-6}=\frac{-26+2\sqrt{139}}{-12}
$