(4u+5)(2+u)=0

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

(4u+5)(2+u)=0

We add all the numbers together, and all the variables

(4u+5)(u+2)=0

We multiply parentheses ..

(+4u^2+8u+5u+10)=0

We get rid of parentheses

4u^2+8u+5u+10=0

We add all the numbers together, and all the variables

4u^2+13u+10=0

a = 4; b = 13; c = +10;
Δ = b2-4ac
Δ = 132-4·4·10
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*4}=\frac{-16}{8}
=-2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*4}=\frac{-10}{8}
=-1+1/4
$