1200=(2x-20)(5x-20)

Solution for 1200=(2x-20)(5x-20) equation:

1200=(2x-20)(5x-20)

We move all terms to the left:

1200-((2x-20)(5x-20))=0

We multiply parentheses ..

-((+10x^2-40x-100x+400))+1200=0


We calculate terms in parentheses: -((+10x^2-40x-100x+400)), so:

(+10x^2-40x-100x+400)

We get rid of parentheses

10x^2-40x-100x+400

We add all the numbers together, and all the variables

10x^2-140x+400

Back to the equation:

-(10x^2-140x+400)


We get rid of parentheses

-10x^2+140x-400+1200=0

We add all the numbers together, and all the variables

-10x^2+140x+800=0

a = -10; b = 140; c = +800;
Δ = b2-4ac
Δ = 1402-4·(-10)·800
Δ = 51600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{51600}=\sqrt{400*129}=\sqrt{400}*\sqrt{129}=20\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(140)-20\sqrt{129}}{2*-10}=\frac{-140-20\sqrt{129}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(140)+20\sqrt{129}}{2*-10}=\frac{-140+20\sqrt{129}}{-20}
$