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13x^2+20x-92=0
a = 13; b = 20; c = -92;
Δ = b2-4ac
Δ = 202-4·13·(-92)
Δ = 5184
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{5184}=72$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-72}{2*13}=\frac{-92}{26} =-3+7/13 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+72}{2*13}=\frac{52}{26} =2 $
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