(3-c)(4-c)=2c^2-20c+48

Solution for (3-c)(4-c)=2c^2-20c+48 equation:

(3-c)(4-c)=2c^2-20c+48

We move all terms to the left:

(3-c)(4-c)-(2c^2-20c+48)=0

We add all the numbers together, and all the variables

(-1c+3)(-1c+4)-(2c^2-20c+48)=0

We get rid of parentheses

-2c^2+(-1c+3)(-1c+4)+20c-48=0

We multiply parentheses ..

-2c^2+(+c^2-4c-3c+12)+20c-48=0

We get rid of parentheses

-2c^2+c^2-4c-3c+20c+12-48=0

We add all the numbers together, and all the variables

-1c^2+13c-36=0

a = -1; b = 13; c = -36;
Δ = b2-4ac
Δ = 132-4·(-1)·(-36)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*-1}=\frac{-18}{-2}
=+9
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*-1}=\frac{-8}{-2}
=+4
$