(4u-5)(2u-7)=0

Solution for (4u-5)(2u-7)=0 equation:

(4u-5)(2u-7)=0

We multiply parentheses ..

(+8u^2-28u-10u+35)=0

We get rid of parentheses

8u^2-28u-10u+35=0

We add all the numbers together, and all the variables

8u^2-38u+35=0

a = 8; b = -38; c = +35;
Δ = b2-4ac
Δ = -382-4·8·35
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-18}{2*8}=\frac{20}{16}
=1+1/4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+18}{2*8}=\frac{56}{16}
=3+1/2
$