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4x^2=8
We move all terms to the left:
4x^2-(8)=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} =-\sqrt{2} $$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} =\sqrt{2} $
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