(4+u)(5u+7)=0

Solution for (4+u)(5u+7)=0 equation:

(4+u)(5u+7)=0

We add all the numbers together, and all the variables

(u+4)(5u+7)=0

We multiply parentheses ..

(+5u^2+7u+20u+28)=0

We get rid of parentheses

5u^2+7u+20u+28=0

We add all the numbers together, and all the variables

5u^2+27u+28=0

a = 5; b = 27; c = +28;
Δ = b2-4ac
Δ = 272-4·5·28
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-13}{2*5}=\frac{-40}{10}
=-4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+13}{2*5}=\frac{-14}{10}
=-1+2/5
$