(5c+8)(c-4)=0

Solution for (5c+8)(c-4)=0 equation:

(5c+8)(c-4)=0

We multiply parentheses ..

(+5c^2-20c+8c-32)=0

We get rid of parentheses

5c^2-20c+8c-32=0

We add all the numbers together, and all the variables

5c^2-12c-32=0

a = 5; b = -12; c = -32;
Δ = b2-4ac
Δ = -122-4·5·(-32)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-28}{2*5}=\frac{-16}{10}
=-1+3/5
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+28}{2*5}=\frac{40}{10}
=4
$