(4c-5)(7c+5)=0

Solution for (4c-5)(7c+5)=0 equation:

(4c-5)(7c+5)=0

We multiply parentheses ..

(+28c^2+20c-35c-25)=0

We get rid of parentheses

28c^2+20c-35c-25=0

We add all the numbers together, and all the variables

28c^2-15c-25=0

a = 28; b = -15; c = -25;
Δ = b2-4ac
Δ = -152-4·28·(-25)
Δ = 3025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3025}=55$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-55}{2*28}=\frac{-40}{56}
=-5/7
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+55}{2*28}=\frac{70}{56}
=1+1/4
$