(3u-5)(2+u)=0

Solution for (3u-5)(2+u)=0 equation:

(3u-5)(2+u)=0

We add all the numbers together, and all the variables

(3u-5)(u+2)=0

We multiply parentheses ..

(+3u^2+6u-5u-10)=0

We get rid of parentheses

3u^2+6u-5u-10=0

We add all the numbers together, and all the variables

3u^2+u-10=0

a = 3; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·3·(-10)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*3}=\frac{-12}{6}
=-2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*3}=\frac{10}{6}
=1+2/3
$