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13x^2+12x-240=0
a = 13; b = 12; c = -240;
Δ = b2-4ac
Δ = 122-4·13·(-240)
Δ = 12624
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12624}=\sqrt{16*789}=\sqrt{16}*\sqrt{789}=4\sqrt{789}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{789}}{2*13}=\frac{-12-4\sqrt{789}}{26} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{789}}{2*13}=\frac{-12+4\sqrt{789}}{26} $
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