(1+u)(4u+6)=0

Solution for (1+u)(4u+6)=0 equation:

(1+u)(4u+6)=0

We add all the numbers together, and all the variables

(u+1)(4u+6)=0

We multiply parentheses ..

(+4u^2+6u+4u+6)=0

We get rid of parentheses

4u^2+6u+4u+6=0

We add all the numbers together, and all the variables

4u^2+10u+6=0

a = 4; b = 10; c = +6;
Δ = b2-4ac
Δ = 102-4·4·6
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*4}=\frac{-12}{8}
=-1+1/2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*4}=\frac{-8}{8}
=-1
$