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

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

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

We add all the numbers together, and all the variables

(-1u+2)(4u-7)=0

We multiply parentheses ..

(-4u^2+7u+8u-14)=0

We get rid of parentheses

-4u^2+7u+8u-14=0

We add all the numbers together, and all the variables

-4u^2+15u-14=0

a = -4; b = 15; c = -14;
Δ = b2-4ac
Δ = 152-4·(-4)·(-14)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-1}{2*-4}=\frac{-16}{-8}
=+2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+1}{2*-4}=\frac{-14}{-8}
=1+3/4
$