(2x+9)(6x+12)=16x^2+78x+44

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Solution for (2x+9)(6x+12)=16x^2+78x+44 equation:



(2x+9)(6x+12)=16x^2+78x+44
We move all terms to the left:
(2x+9)(6x+12)-(16x^2+78x+44)=0
We get rid of parentheses
-16x^2+(2x+9)(6x+12)-78x-44=0
We multiply parentheses ..
-16x^2+(+12x^2+24x+54x+108)-78x-44=0
We get rid of parentheses
-16x^2+12x^2+24x+54x-78x+108-44=0
We add all the numbers together, and all the variables
-4x^2+64=0
a = -4; b = 0; c = +64;
Δ = b2-4ac
Δ = 02-4·(-4)·64
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*-4}=\frac{-32}{-8} =+4 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*-4}=\frac{32}{-8} =-4 $

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