(3x-4)(x-5)=14

Solution for (3x-4)(x-5)=14 equation:

(3x-4)(x-5)=14

We move all terms to the left:

(3x-4)(x-5)-(14)=0

We multiply parentheses ..

(+3x^2-15x-4x+20)-14=0

We get rid of parentheses

3x^2-15x-4x+20-14=0

We add all the numbers together, and all the variables

3x^2-19x+6=0

a = 3; b = -19; c = +6;
Δ = b2-4ac
Δ = -192-4·3·6
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-17}{2*3}=\frac{2}{6}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+17}{2*3}=\frac{36}{6}
=6
$