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13x^2+5x-300=0
a = 13; b = 5; c = -300;
Δ = b2-4ac
Δ = 52-4·13·(-300)
Δ = 15625
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{15625}=125$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-125}{2*13}=\frac{-130}{26} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+125}{2*13}=\frac{120}{26} =4+8/13 $
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