(5x^2-40x)-(2x+16)=0

Solution for (5x^2-40x)-(2x+16)=0 equation:

(5x^2-40x)-(2x+16)=0

We get rid of parentheses

5x^2-40x-2x-16=0

We add all the numbers together, and all the variables

5x^2-42x-16=0

a = 5; b = -42; c = -16;
Δ = b2-4ac
Δ = -422-4·5·(-16)
Δ = 2084
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:

$\sqrt{\Delta}=\sqrt{2084}=\sqrt{4*521}=\sqrt{4}*\sqrt{521}=2\sqrt{521}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{521}}{2*5}=\frac{42-2\sqrt{521}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{521}}{2*5}=\frac{42+2\sqrt{521}}{10}
$