(x-20)(2x+10)=(x+40)

Solution for (x-20)(2x+10)=(x+40) equation:

(x-20)(2x+10)=(x+40)

We move all terms to the left:

(x-20)(2x+10)-((x+40))=0

We multiply parentheses ..

(+2x^2+10x-40x-200)-((x+40))=0


We calculate terms in parentheses: -((x+40)), so:

(x+40)

We get rid of parentheses

x+40

Back to the equation:

-(x+40)


We get rid of parentheses

2x^2+10x-40x-x-200-40=0

We add all the numbers together, and all the variables

2x^2-31x-240=0

a = 2; b = -31; c = -240;
Δ = b2-4ac
Δ = -312-4·2·(-240)
Δ = 2881
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{-(-31)-\sqrt{2881}}{2*2}=\frac{31-\sqrt{2881}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{2881}}{2*2}=\frac{31+\sqrt{2881}}{4}
$