x=(x+40)(2x-22)

Solution for x=(x+40)(2x-22) equation:

x=(x+40)(2x-22)

We move all terms to the left:

x-((x+40)(2x-22))=0

We multiply parentheses ..

-((+2x^2-22x+80x-880))+x=0


We calculate terms in parentheses: -((+2x^2-22x+80x-880)), so:

(+2x^2-22x+80x-880)

We get rid of parentheses

2x^2-22x+80x-880

We add all the numbers together, and all the variables

2x^2+58x-880

Back to the equation:

-(2x^2+58x-880)


We add all the numbers together, and all the variables

x-(2x^2+58x-880)=0

We get rid of parentheses

-2x^2+x-58x+880=0

We add all the numbers together, and all the variables

-2x^2-57x+880=0

a = -2; b = -57; c = +880;
Δ = b2-4ac
Δ = -572-4·(-2)·880
Δ = 10289
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{-(-57)-\sqrt{10289}}{2*-2}=\frac{57-\sqrt{10289}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-57)+\sqrt{10289}}{2*-2}=\frac{57+\sqrt{10289}}{-4}
$