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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4u^2-1u+24u-6)=0

We get rid of parentheses

4u^2-1u+24u-6=0

We add all the numbers together, and all the variables

4u^2+23u-6=0

a = 4; b = 23; c = -6;
Δ = b2-4ac
Δ = 232-4·4·(-6)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-25}{2*4}=\frac{-48}{8}
=-6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+25}{2*4}=\frac{2}{8}
=1/4
$