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

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

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

We add all the numbers together, and all the variables

(5s+3)(-1s+2)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-5s^2+7s+6=0

a = -5; b = 7; c = +6;
Δ = b2-4ac
Δ = 72-4·(-5)·6
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*-5}=\frac{-20}{-10}
=+2
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*-5}=\frac{6}{-10}
=-3/5
$