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5=26x^2
We move all terms to the left:
5-(26x^2)=0
a = -26; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-26)·5
Δ = 520
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{130}}{2*-26}=\frac{0-2\sqrt{130}}{-52} =-\frac{2\sqrt{130}}{-52} =-\frac{\sqrt{130}}{-26} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{130}}{2*-26}=\frac{0+2\sqrt{130}}{-52} =\frac{2\sqrt{130}}{-52} =\frac{\sqrt{130}}{-26} $
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