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8=4x^2
We move all terms to the left:
8-(4x^2)=0
a = -4; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-4)·8
Δ = 128
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*-4}=\frac{0-8\sqrt{2}}{-8} =-\frac{8\sqrt{2}}{-8} =-\frac{\sqrt{2}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*-4}=\frac{0+8\sqrt{2}}{-8} =\frac{8\sqrt{2}}{-8} =\frac{\sqrt{2}}{-1} $
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