(8x-20)(10x-11)=43

Solution for (8x-20)(10x-11)=43 equation:

(8x-20)(10x-11)=43

We move all terms to the left:

(8x-20)(10x-11)-(43)=0

We multiply parentheses ..

(+80x^2-88x-200x+220)-43=0

We get rid of parentheses

80x^2-88x-200x+220-43=0

We add all the numbers together, and all the variables

80x^2-288x+177=0

a = 80; b = -288; c = +177;
Δ = b2-4ac
Δ = -2882-4·80·177
Δ = 26304
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:

$\sqrt{\Delta}=\sqrt{26304}=\sqrt{64*411}=\sqrt{64}*\sqrt{411}=8\sqrt{411}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-288)-8\sqrt{411}}{2*80}=\frac{288-8\sqrt{411}}{160}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-288)+8\sqrt{411}}{2*80}=\frac{288+8\sqrt{411}}{160}
$