X=(x+5)(2x-8)

Solution for X=(x+5)(2x-8) equation:

X=(X+5)(2X-8)

We move all terms to the left:

X-((X+5)(2X-8))=0

We multiply parentheses ..

-((+2X^2-8X+10X-40))+X=0


We calculate terms in parentheses: -((+2X^2-8X+10X-40)), so:

(+2X^2-8X+10X-40)

We get rid of parentheses

2X^2-8X+10X-40

We add all the numbers together, and all the variables

2X^2+2X-40

Back to the equation:

-(2X^2+2X-40)


We add all the numbers together, and all the variables

X-(2X^2+2X-40)=0

We get rid of parentheses

-2X^2+X-2X+40=0

We add all the numbers together, and all the variables

-2X^2-1X+40=0

a = -2; b = -1; c = +40;
Δ = b2-4ac
Δ = -12-4·(-2)·40
Δ = 321
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{-(-1)-\sqrt{321}}{2*-2}=\frac{1-\sqrt{321}}{-4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{321}}{2*-2}=\frac{1+\sqrt{321}}{-4}
$