X=(3x-10)(x+40)

Solution for X=(3x-10)(x+40) equation:

X=(3X-10)(X+40)

We move all terms to the left:

X-((3X-10)(X+40))=0

We multiply parentheses ..

-((+3X^2+120X-10X-400))+X=0


We calculate terms in parentheses: -((+3X^2+120X-10X-400)), so:

(+3X^2+120X-10X-400)

We get rid of parentheses

3X^2+120X-10X-400

We add all the numbers together, and all the variables

3X^2+110X-400

Back to the equation:

-(3X^2+110X-400)


We add all the numbers together, and all the variables

X-(3X^2+110X-400)=0

We get rid of parentheses

-3X^2+X-110X+400=0

We add all the numbers together, and all the variables

-3X^2-109X+400=0

a = -3; b = -109; c = +400;
Δ = b2-4ac
Δ = -1092-4·(-3)·400
Δ = 16681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-109)-\sqrt{16681}}{2*-3}=\frac{109-\sqrt{16681}}{-6}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-109)+\sqrt{16681}}{2*-3}=\frac{109+\sqrt{16681}}{-6}
$