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26x^2+35x-36=0
a = 26; b = 35; c = -36;
Δ = b2-4ac
Δ = 352-4·26·(-36)
Δ = 4969
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-\sqrt{4969}}{2*26}=\frac{-35-\sqrt{4969}}{52} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+\sqrt{4969}}{2*26}=\frac{-35+\sqrt{4969}}{52} $
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