(4u+3)(5+u)=0

Solution for (4u+3)(5+u)=0 equation:

(4u+3)(5+u)=0

We add all the numbers together, and all the variables

(4u+3)(u+5)=0

We multiply parentheses ..

(+4u^2+20u+3u+15)=0

We get rid of parentheses

4u^2+20u+3u+15=0

We add all the numbers together, and all the variables

4u^2+23u+15=0

a = 4; b = 23; c = +15;
Δ = b2-4ac
Δ = 232-4·4·15
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*4}=\frac{-40}{8}
=-5
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*4}=\frac{-6}{8}
=-3/4
$