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2x^2=1040
We move all terms to the left:
2x^2-(1040)=0
a = 2; b = 0; c = -1040;
Δ = b2-4ac
Δ = 02-4·2·(-1040)
Δ = 8320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8320}=\sqrt{64*130}=\sqrt{64}*\sqrt{130}=8\sqrt{130}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{130}}{2*2}=\frac{0-8\sqrt{130}}{4} =-\frac{8\sqrt{130}}{4} =-2\sqrt{130} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{130}}{2*2}=\frac{0+8\sqrt{130}}{4} =\frac{8\sqrt{130}}{4} =2\sqrt{130} $
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