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20(2^2x+1)=500
We move all terms to the left:
20(2^2x+1)-(500)=0
We multiply parentheses
40x^2+20-500=0
We add all the numbers together, and all the variables
40x^2-480=0
a = 40; b = 0; c = -480;
Δ = b2-4ac
Δ = 02-4·40·(-480)
Δ = 76800
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{76800}=\sqrt{25600*3}=\sqrt{25600}*\sqrt{3}=160\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-160\sqrt{3}}{2*40}=\frac{0-160\sqrt{3}}{80} =-\frac{160\sqrt{3}}{80} =-2\sqrt{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+160\sqrt{3}}{2*40}=\frac{0+160\sqrt{3}}{80} =\frac{160\sqrt{3}}{80} =2\sqrt{3} $
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