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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+8x^2-6x-20x+15)=0

We get rid of parentheses

8x^2-6x-20x+15=0

We add all the numbers together, and all the variables

8x^2-26x+15=0

a = 8; b = -26; c = +15;
Δ = b2-4ac
Δ = -262-4·8·15
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-14}{2*8}=\frac{12}{16}
=3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+14}{2*8}=\frac{40}{16}
=2+1/2
$