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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-2u^2+4u-5u+10)=0

We get rid of parentheses

-2u^2+4u-5u+10=0

We add all the numbers together, and all the variables

-2u^2-1u+10=0

a = -2; b = -1; c = +10;
Δ = b2-4ac
Δ = -12-4·(-2)·10
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-9}{2*-2}=\frac{-8}{-4}
=+2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+9}{2*-2}=\frac{10}{-4}
=-2+1/2
$