(4c+5)(5c+1)=

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

(4c+5)(5c+1)=

We move all terms to the left:

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

We add all the numbers together, and all the variables

(4c+5)(5c+1)=0

We multiply parentheses ..

(+20c^2+4c+25c+5)=0

We get rid of parentheses

20c^2+4c+25c+5=0

We add all the numbers together, and all the variables

20c^2+29c+5=0

a = 20; b = 29; c = +5;
Δ = b2-4ac
Δ = 292-4·20·5
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-21}{2*20}=\frac{-50}{40}
=-1+1/4
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+21}{2*20}=\frac{-8}{40}
=-1/5
$