(21x-4)(19x=4)

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Solution for (21x-4)(19x=4) equation:



(21x-4)(19x=4)
We move all terms to the left:
(21x-4)(19x-(4))=0
We multiply parentheses ..
(+399x^2-84x-76x+16)=0
We get rid of parentheses
399x^2-84x-76x+16=0
We add all the numbers together, and all the variables
399x^2-160x+16=0
a = 399; b = -160; c = +16;
Δ = b2-4ac
Δ = -1602-4·399·16
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-8}{2*399}=\frac{152}{798} =4/21 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+8}{2*399}=\frac{168}{798} =4/19 $

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