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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

-((3x+5)(-1x+2))+0=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 calculate terms in parentheses: -((-3x^2+6x-5x+10)), so:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2+x+10

Back to the equation:

-(-3x^2+x+10)


We get rid of parentheses

3x^2-x-10=0

We add all the numbers together, and all the variables

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