0=4u^2+19u-28

Solution for 0=4u^2+19u-28 equation:

0=4u^2+19u-28

We move all terms to the left:

0-(4u^2+19u-28)=0

We add all the numbers together, and all the variables

-(4u^2+19u-28)=0

We get rid of parentheses

-4u^2-19u+28=0

a = -4; b = -19; c = +28;
Δ = b2-4ac
Δ = -192-4·(-4)·28
Δ = 809
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{809}}{2*-4}=\frac{19-\sqrt{809}}{-8}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{809}}{2*-4}=\frac{19+\sqrt{809}}{-8}
$