(1+u)(2u-5)=0

Solution for (1+u)(2u-5)=0 equation:

(1+u)(2u-5)=0

We add all the numbers together, and all the variables

(u+1)(2u-5)=0

We multiply parentheses ..

(+2u^2-5u+2u-5)=0

We get rid of parentheses

2u^2-5u+2u-5=0

We add all the numbers together, and all the variables

2u^2-3u-5=0

a = 2; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·2·(-5)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*2}=\frac{-4}{4}
=-1
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*2}=\frac{10}{4}
=2+1/2
$