(4u+9)(1+u)=0

Solution for (4u+9)(1+u)=0 equation:

(4u+9)(1+u)=0

We add all the numbers together, and all the variables

(4u+9)(u+1)=0

We multiply parentheses ..

(+4u^2+4u+9u+9)=0

We get rid of parentheses

4u^2+4u+9u+9=0

We add all the numbers together, and all the variables

4u^2+13u+9=0

a = 4; b = 13; c = +9;
Δ = b2-4ac
Δ = 132-4·4·9
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*4}=\frac{-18}{8}
=-2+1/4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*4}=\frac{-8}{8}
=-1
$