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2x^2+4=340
We move all terms to the left:
2x^2+4-(340)=0
We add all the numbers together, and all the variables
2x^2-336=0
a = 2; b = 0; c = -336;
Δ = b2-4ac
Δ = 02-4·2·(-336)
Δ = 2688
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2688}=\sqrt{64*42}=\sqrt{64}*\sqrt{42}=8\sqrt{42}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{42}}{2*2}=\frac{0-8\sqrt{42}}{4} =-\frac{8\sqrt{42}}{4} =-2\sqrt{42} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{42}}{2*2}=\frac{0+8\sqrt{42}}{4} =\frac{8\sqrt{42}}{4} =2\sqrt{42} $
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