(2c+5)(7-2c)+(2c+5)(8c+9)=0

Solution for (2c+5)(7-2c)+(2c+5)(8c+9)=0 equation:

(2c+5)(7-2c)+(2c+5)(8c+9)=0

We add all the numbers together, and all the variables

(2c+5)(-2c+7)+(2c+5)(8c+9)=0

We multiply parentheses ..

(-4c^2+14c-10c+35)+(2c+5)(8c+9)=0

We get rid of parentheses

-4c^2+14c-10c+(2c+5)(8c+9)+35=0

We multiply parentheses ..

-4c^2+(+16c^2+18c+40c+45)+14c-10c+35=0

We add all the numbers together, and all the variables

-4c^2+(+16c^2+18c+40c+45)+4c+35=0

We get rid of parentheses

-4c^2+16c^2+18c+40c+4c+45+35=0

We add all the numbers together, and all the variables

12c^2+62c+80=0

a = 12; b = 62; c = +80;
Δ = b2-4ac
Δ = 622-4·12·80
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-2}{2*12}=\frac{-64}{24}
=-2+2/3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+2}{2*12}=\frac{-60}{24}
=-2+1/2
$