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16x^2-13x-16=0
a = 16; b = -13; c = -16;
Δ = b2-4ac
Δ = -132-4·16·(-16)
Δ = 1193
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{-(-13)-\sqrt{1193}}{2*16}=\frac{13-\sqrt{1193}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{1193}}{2*16}=\frac{13+\sqrt{1193}}{32} $
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