(7c+5)(2-9c)=0

Solution for (7c+5)(2-9c)=0 equation:

(7c+5)(2-9c)=0

We add all the numbers together, and all the variables

(7c+5)(-9c+2)=0

We multiply parentheses ..

(-63c^2+14c-45c+10)=0

We get rid of parentheses

-63c^2+14c-45c+10=0

We add all the numbers together, and all the variables

-63c^2-31c+10=0

a = -63; b = -31; c = +10;
Δ = b2-4ac
Δ = -312-4·(-63)·10
Δ = 3481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3481}=59$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-59}{2*-63}=\frac{-28}{-126}
=2/9
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+59}{2*-63}=\frac{90}{-126}
=-5/7
$