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

Solution for (2u+5)(u+1)=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

2u^2+2u+5u+5=0

We add all the numbers together, and all the variables

2u^2+7u+5=0

a = 2; b = 7; c = +5;
Δ = b2-4ac
Δ = 72-4·2·5
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{-(7)-3}{2*2}=\frac{-10}{4}
=-2+1/2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3}{2*2}=\frac{-4}{4}
=-1
$