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7x^2-13=0
a = 7; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·7·(-13)
Δ = 364
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{364}=\sqrt{4*91}=\sqrt{4}*\sqrt{91}=2\sqrt{91}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{91}}{2*7}=\frac{0-2\sqrt{91}}{14} =-\frac{2\sqrt{91}}{14} =-\frac{\sqrt{91}}{7} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{91}}{2*7}=\frac{0+2\sqrt{91}}{14} =\frac{2\sqrt{91}}{14} =\frac{\sqrt{91}}{7} $
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