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

Solution for (7+u)(4u+9)=0 equation:

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

We add all the numbers together, and all the variables

(u+7)(4u+9)=0

We multiply parentheses ..

(+4u^2+9u+28u+63)=0

We get rid of parentheses

4u^2+9u+28u+63=0

We add all the numbers together, and all the variables

4u^2+37u+63=0

a = 4; b = 37; c = +63;
Δ = b2-4ac
Δ = 372-4·4·63
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(37)-19}{2*4}=\frac{-56}{8}
=-7
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+19}{2*4}=\frac{-18}{8}
=-2+1/4
$