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16x^2-13x-2=0
a = 16; b = -13; c = -2;
Δ = b2-4ac
Δ = -132-4·16·(-2)
Δ = 297
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3\sqrt{33}}{2*16}=\frac{13-3\sqrt{33}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3\sqrt{33}}{2*16}=\frac{13+3\sqrt{33}}{32} $
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