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(3x-4)(4x-3)-(2x-7)(3x-2)=124
We move all terms to the left:
(3x-4)(4x-3)-(2x-7)(3x-2)-(124)=0
We multiply parentheses ..
(+12x^2-9x-16x+12)-(2x-7)(3x-2)-124=0
We get rid of parentheses
12x^2-9x-16x-(2x-7)(3x-2)+12-124=0
We multiply parentheses ..
12x^2-(+6x^2-4x-21x+14)-9x-16x+12-124=0
We add all the numbers together, and all the variables
12x^2-(+6x^2-4x-21x+14)-25x-112=0
We get rid of parentheses
12x^2-6x^2+4x+21x-25x-14-112=0
We add all the numbers together, and all the variables
6x^2-126=0
a = 6; b = 0; c = -126;
Δ = b2-4ac
Δ = 02-4·6·(-126)
Δ = 3024
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3024}=\sqrt{144*21}=\sqrt{144}*\sqrt{21}=12\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{21}}{2*6}=\frac{0-12\sqrt{21}}{12} =-\frac{12\sqrt{21}}{12} =-\sqrt{21} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{21}}{2*6}=\frac{0+12\sqrt{21}}{12} =\frac{12\sqrt{21}}{12} =\sqrt{21} $
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