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(12x)2-3x=7x+8-10x
We move all terms to the left:
(12x)2-3x-(7x+8-10x)=0
We add all the numbers together, and all the variables
12x2-3x-(-3x+8)=0
We add all the numbers together, and all the variables
12x^2-3x-(-3x+8)=0
We get rid of parentheses
12x^2-3x+3x-8=0
We add all the numbers together, and all the variables
12x^2-8=0
a = 12; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·12·(-8)
Δ = 384
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*12}=\frac{0-8\sqrt{6}}{24} =-\frac{8\sqrt{6}}{24} =-\frac{\sqrt{6}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*12}=\frac{0+8\sqrt{6}}{24} =\frac{8\sqrt{6}}{24} =\frac{\sqrt{6}}{3} $
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