(4x-1)(4x-1)=(x+1)(10x+5)

Solution for (4x-1)(4x-1)=(x+1)(10x+5) equation:

(4x-1)(4x-1)=(x+1)(10x+5)

We move all terms to the left:

(4x-1)(4x-1)-((x+1)(10x+5))=0

We multiply parentheses ..

(+16x^2-4x-4x+1)-((x+1)(10x+5))=0


We calculate terms in parentheses: -((x+1)(10x+5)), so:

(x+1)(10x+5)

We multiply parentheses ..

(+10x^2+5x+10x+5)

We get rid of parentheses

10x^2+5x+10x+5

We add all the numbers together, and all the variables

10x^2+15x+5

Back to the equation:

-(10x^2+15x+5)


We get rid of parentheses

16x^2-10x^2-4x-4x-15x+1-5=0

We add all the numbers together, and all the variables

6x^2-23x-4=0

a = 6; b = -23; c = -4;
Δ = b2-4ac
Δ = -232-4·6·(-4)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-25}{2*6}=\frac{-2}{12}
=-1/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+25}{2*6}=\frac{48}{12}
=4
$