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5x^2-4=10x^2-20
We move all terms to the left:
5x^2-4-(10x^2-20)=0
We get rid of parentheses
5x^2-10x^2+20-4=0
We add all the numbers together, and all the variables
-5x^2+16=0
a = -5; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-5)·16
Δ = 320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*-5}=\frac{0-8\sqrt{5}}{-10} =-\frac{8\sqrt{5}}{-10} =-\frac{4\sqrt{5}}{-5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*-5}=\frac{0+8\sqrt{5}}{-10} =\frac{8\sqrt{5}}{-10} =\frac{4\sqrt{5}}{-5} $
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