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40*32=(3x-6)(x+6)
We move all terms to the left:
40*32-((3x-6)(x+6))=0
We add all the numbers together, and all the variables
-((3x-6)(x+6))+1280=0
We multiply parentheses ..
-((+3x^2+18x-6x-36))+1280=0
We calculate terms in parentheses: -((+3x^2+18x-6x-36)), so:We get rid of parentheses
(+3x^2+18x-6x-36)
We get rid of parentheses
3x^2+18x-6x-36
We add all the numbers together, and all the variables
3x^2+12x-36
Back to the equation:
-(3x^2+12x-36)
-3x^2-12x+36+1280=0
We add all the numbers together, and all the variables
-3x^2-12x+1316=0
a = -3; b = -12; c = +1316;
Δ = b2-4ac
Δ = -122-4·(-3)·1316
Δ = 15936
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{15936}=\sqrt{64*249}=\sqrt{64}*\sqrt{249}=8\sqrt{249}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{249}}{2*-3}=\frac{12-8\sqrt{249}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{249}}{2*-3}=\frac{12+8\sqrt{249}}{-6} $
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