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13x^2+26x-39=0
a = 13; b = 26; c = -39;
Δ = b2-4ac
Δ = 262-4·13·(-39)
Δ = 2704
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{2704}=52$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-52}{2*13}=\frac{-78}{26} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+52}{2*13}=\frac{26}{26} =1 $
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