5c(4c-1)=18c+21

Solution for 5c(4c-1)=18c+21 equation:

5c(4c-1)=18c+21

We move all terms to the left:

5c(4c-1)-(18c+21)=0

We multiply parentheses

20c^2-5c-(18c+21)=0

We get rid of parentheses

20c^2-5c-18c-21=0

We add all the numbers together, and all the variables

20c^2-23c-21=0

a = 20; b = -23; c = -21;
Δ = b2-4ac
Δ = -232-4·20·(-21)
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-47}{2*20}=\frac{-24}{40}
=-3/5
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+47}{2*20}=\frac{70}{40}
=1+3/4
$