(2f-5)(4f-9)=0

Solution for (2f-5)(4f-9)=0 equation:

(2f-5)(4f-9)=0

We multiply parentheses ..

(+8f^2-18f-20f+45)=0

We get rid of parentheses

8f^2-18f-20f+45=0

We add all the numbers together, and all the variables

8f^2-38f+45=0

a = 8; b = -38; c = +45;
Δ = b2-4ac
Δ = -382-4·8·45
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2}{2*8}=\frac{36}{16}
=2+1/4
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2}{2*8}=\frac{40}{16}
=2+1/2
$