19x=(2x-19)(2x-19)

Solution for 19x=(2x-19)(2x-19) equation:

19x=(2x-19)(2x-19)

We move all terms to the left:

19x-((2x-19)(2x-19))=0

We multiply parentheses ..

-((+4x^2-38x-38x+361))+19x=0


We calculate terms in parentheses: -((+4x^2-38x-38x+361)), so:

(+4x^2-38x-38x+361)

We get rid of parentheses

4x^2-38x-38x+361

We add all the numbers together, and all the variables

4x^2-76x+361

Back to the equation:

-(4x^2-76x+361)


We add all the numbers together, and all the variables

19x-(4x^2-76x+361)=0

We get rid of parentheses

-4x^2+19x+76x-361=0

We add all the numbers together, and all the variables

-4x^2+95x-361=0

a = -4; b = 95; c = -361;
Δ = b2-4ac
Δ = 952-4·(-4)·(-361)
Δ = 3249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3249}=57$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(95)-57}{2*-4}=\frac{-152}{-8}
=+19
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(95)+57}{2*-4}=\frac{-38}{-8}
=4+3/4
$