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13x^2-18x+5=0
a = 13; b = -18; c = +5;
Δ = b2-4ac
Δ = -182-4·13·5
Δ = 64
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-8}{2*13}=\frac{10}{26} =5/13 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+8}{2*13}=\frac{26}{26} =1 $
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