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2x^2=5
We move all terms to the left:
2x^2-(5)=0
a = 2; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·2·(-5)
Δ = 40
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*2}=\frac{0-2\sqrt{10}}{4} =-\frac{2\sqrt{10}}{4} =-\frac{\sqrt{10}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*2}=\frac{0+2\sqrt{10}}{4} =\frac{2\sqrt{10}}{4} =\frac{\sqrt{10}}{2} $
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