22(20-x)+36(20+x)=3(20+x)(20-x)

Solution for 22(20-x)+36(20+x)=3(20+x)(20-x) equation:

22(20-x)+36(20+x)=3(20+x)(20-x)

We move all terms to the left:

22(20-x)+36(20+x)-(3(20+x)(20-x))=0

We add all the numbers together, and all the variables

22(-1x+20)+36(x+20)-(3(x+20)(-1x+20))=0

We multiply parentheses

-22x+36x-(3(x+20)(-1x+20))+440+720=0

We multiply parentheses ..

-(3(-1x^2+20x-20x+400))-22x+36x+440+720=0


We calculate terms in parentheses: -(3(-1x^2+20x-20x+400)), so:

3(-1x^2+20x-20x+400)

We multiply parentheses

-3x^2+60x-60x+1200

We add all the numbers together, and all the variables

-3x^2+1200

Back to the equation:

-(-3x^2+1200)


We add all the numbers together, and all the variables

-(-3x^2+1200)+14x+1160=0

We get rid of parentheses

3x^2+14x-1200+1160=0

We add all the numbers together, and all the variables

3x^2+14x-40=0

a = 3; b = 14; c = -40;
Δ = b2-4ac
Δ = 142-4·3·(-40)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-26}{2*3}=\frac{-40}{6}
=-6+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+26}{2*3}=\frac{12}{6}
=2
$