(4c-3)(c=5)

Solution for (4c-3)(c=5) equation:

(4c-3)(c=5)

We move all terms to the left:

(4c-3)(c-(5))=0

We multiply parentheses ..

(+4c^2-20c-3c+15)=0

We get rid of parentheses

4c^2-20c-3c+15=0

We add all the numbers together, and all the variables

4c^2-23c+15=0

a = 4; b = -23; c = +15;
Δ = b2-4ac
Δ = -232-4·4·15
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-17}{2*4}=\frac{6}{8}
=3/4
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+17}{2*4}=\frac{40}{8}
=5
$