(3+u)(4u-5)=0

Solution for (3+u)(4u-5)=0 equation:

(3+u)(4u-5)=0

We add all the numbers together, and all the variables

(u+3)(4u-5)=0

We multiply parentheses ..

(+4u^2-5u+12u-15)=0

We get rid of parentheses

4u^2-5u+12u-15=0

We add all the numbers together, and all the variables

4u^2+7u-15=0

a = 4; b = 7; c = -15;
Δ = b2-4ac
Δ = 72-4·4·(-15)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*4}=\frac{-24}{8}
=-3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*4}=\frac{10}{8}
=1+1/4
$