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

Solution for 0=(5x-3)(x-4) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+5x^2-20x-3x+12))=0


We calculate terms in parentheses: -((+5x^2-20x-3x+12)), so:

(+5x^2-20x-3x+12)

We get rid of parentheses

5x^2-20x-3x+12

We add all the numbers together, and all the variables

5x^2-23x+12

Back to the equation:

-(5x^2-23x+12)


We get rid of parentheses

-5x^2+23x-12=0

a = -5; b = 23; c = -12;
Δ = b2-4ac
Δ = 232-4·(-5)·(-12)
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{-(23)-17}{2*-5}=\frac{-40}{-10}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*-5}=\frac{-6}{-10}
=3/5
$