(3c-5)(c=3)

Solution for (3c-5)(c=3) equation:

(3c-5)(c=3)

We move all terms to the left:

(3c-5)(c-(3))=0

We multiply parentheses ..

(+3c^2-9c-5c+15)=0

We get rid of parentheses

3c^2-9c-5c+15=0

We add all the numbers together, and all the variables

3c^2-14c+15=0

a = 3; b = -14; c = +15;
Δ = b2-4ac
Δ = -142-4·3·15
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4}{2*3}=\frac{10}{6}
=1+2/3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4}{2*3}=\frac{18}{6}
=3
$