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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4u^2+19u-63=0

a = 4; b = 19; c = -63;
Δ = b2-4ac
Δ = 192-4·4·(-63)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-37}{2*4}=\frac{-56}{8}
=-7
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+37}{2*4}=\frac{18}{8}
=2+1/4
$