(2c-7)(-3c+5)=0

Solution for (2c-7)(-3c+5)=0 equation:

(2c-7)(-3c+5)=0

We multiply parentheses ..

(-6c^2+10c+21c-35)=0

We get rid of parentheses

-6c^2+10c+21c-35=0

We add all the numbers together, and all the variables

-6c^2+31c-35=0

a = -6; b = 31; c = -35;
Δ = b2-4ac
Δ = 312-4·(-6)·(-35)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-11}{2*-6}=\frac{-42}{-12}
=3+1/2
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+11}{2*-6}=\frac{-20}{-12}
=1+2/3
$