0=(4x-5)(2x-3)

Solution for 0=(4x-5)(2x-3) equation:

0=(4x-5)(2x-3)

We move all terms to the left:

0-((4x-5)(2x-3))=0

We add all the numbers together, and all the variables

-((4x-5)(2x-3))=0

We multiply parentheses ..

-((+8x^2-12x-10x+15))=0


We calculate terms in parentheses: -((+8x^2-12x-10x+15)), so:

(+8x^2-12x-10x+15)

We get rid of parentheses

8x^2-12x-10x+15

We add all the numbers together, and all the variables

8x^2-22x+15

Back to the equation:

-(8x^2-22x+15)


We get rid of parentheses

-8x^2+22x-15=0

a = -8; b = 22; c = -15;
Δ = b2-4ac
Δ = 222-4·(-8)·(-15)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2}{2*-8}=\frac{-24}{-16}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2}{2*-8}=\frac{-20}{-16}
=1+1/4
$