10=(4x+2)(10x-1)

Solution for 10=(4x+2)(10x-1) equation:

10=(4x+2)(10x-1)

We move all terms to the left:

10-((4x+2)(10x-1))=0

We multiply parentheses ..

-((+40x^2-4x+20x-2))+10=0


We calculate terms in parentheses: -((+40x^2-4x+20x-2)), so:

(+40x^2-4x+20x-2)

We get rid of parentheses

40x^2-4x+20x-2

We add all the numbers together, and all the variables

40x^2+16x-2

Back to the equation:

-(40x^2+16x-2)


We get rid of parentheses

-40x^2-16x+2+10=0

We add all the numbers together, and all the variables

-40x^2-16x+12=0

a = -40; b = -16; c = +12;
Δ = b2-4ac
Δ = -162-4·(-40)·12
Δ = 2176
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:

$\sqrt{\Delta}=\sqrt{2176}=\sqrt{64*34}=\sqrt{64}*\sqrt{34}=8\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{34}}{2*-40}=\frac{16-8\sqrt{34}}{-80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{34}}{2*-40}=\frac{16+8\sqrt{34}}{-80}
$