(4x-7)(4x-5)=0

Solution for (4x-7)(4x-5)=0 equation:

(4x-7)(4x-5)=0

We multiply parentheses ..

(+16x^2-20x-28x+35)=0

We get rid of parentheses

16x^2-20x-28x+35=0

We add all the numbers together, and all the variables

16x^2-48x+35=0

a = 16; b = -48; c = +35;
Δ = b2-4ac
Δ = -482-4·16·35
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8}{2*16}=\frac{40}{32}
=1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8}{2*16}=\frac{56}{32}
=1+3/4
$