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234x^2+(546x^2)-(352x^2)+(1234x^2)=11
We move all terms to the left:
234x^2+(546x^2)-(352x^2)+(1234x^2)-(11)=0
We add all the numbers together, and all the variables
1662x^2-11=0
a = 1662; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·1662·(-11)
Δ = 73128
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{73128}=\sqrt{4*18282}=\sqrt{4}*\sqrt{18282}=2\sqrt{18282}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{18282}}{2*1662}=\frac{0-2\sqrt{18282}}{3324} =-\frac{2\sqrt{18282}}{3324} =-\frac{\sqrt{18282}}{1662} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{18282}}{2*1662}=\frac{0+2\sqrt{18282}}{3324} =\frac{2\sqrt{18282}}{3324} =\frac{\sqrt{18282}}{1662} $
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