2x(3x+4)=19x

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Solution for 2x(3x+4)=19x equation:



2x(3x+4)=19x
We move all terms to the left:
2x(3x+4)-(19x)=0
We add all the numbers together, and all the variables
-19x+2x(3x+4)=0
We multiply parentheses
6x^2-19x+8x=0
We add all the numbers together, and all the variables
6x^2-11x=0
a = 6; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·6·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*6}=\frac{0}{12} =0 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*6}=\frac{22}{12} =1+5/6 $

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