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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2+x+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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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