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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

-4c^2+14c-10c+35=0

We add all the numbers together, and all the variables

-4c^2+4c+35=0

a = -4; b = 4; c = +35;
Δ = b2-4ac
Δ = 42-4·(-4)·35
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-24}{2*-4}=\frac{-28}{-8}
=3+1/2
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+24}{2*-4}=\frac{20}{-8}
=-2+1/2
$