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5x^2-13=0
a = 5; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·5·(-13)
Δ = 260
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{65}}{2*5}=\frac{0-2\sqrt{65}}{10} =-\frac{2\sqrt{65}}{10} =-\frac{\sqrt{65}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{65}}{2*5}=\frac{0+2\sqrt{65}}{10} =\frac{2\sqrt{65}}{10} =\frac{\sqrt{65}}{5} $
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