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13x^2+3x-6=0
a = 13; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·13·(-6)
Δ = 321
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{-(3)-\sqrt{321}}{2*13}=\frac{-3-\sqrt{321}}{26} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{321}}{2*13}=\frac{-3+\sqrt{321}}{26} $
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