(4u+9)(3+u)=0

Solution for (4u+9)(3+u)=0 equation:

(4u+9)(3+u)=0

We add all the numbers together, and all the variables

(4u+9)(u+3)=0

We multiply parentheses ..

(+4u^2+12u+9u+27)=0

We get rid of parentheses

4u^2+12u+9u+27=0

We add all the numbers together, and all the variables

4u^2+21u+27=0

a = 4; b = 21; c = +27;
Δ = b2-4ac
Δ = 212-4·4·27
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{-(21)-3}{2*4}=\frac{-24}{8}
=-3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3}{2*4}=\frac{-18}{8}
=-2+1/4
$