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+60x-400

Back to the equation:

-(10x^2+60x-400)


We get rid of parentheses

-10x^2-60x+400+1200=0

We add all the numbers together, and all the variables

-10x^2-60x+1600=0

a = -10; b = -60; c = +1600;
Δ = b2-4ac
Δ = -602-4·(-10)·1600
Δ = 67600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{67600}=260$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-260}{2*-10}=\frac{-200}{-20}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+260}{2*-10}=\frac{320}{-20}
=-16
$