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16x^2-19x+4=0
a = 16; b = -19; c = +4;
Δ = b2-4ac
Δ = -192-4·16·4
Δ = 105
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{-(-19)-\sqrt{105}}{2*16}=\frac{19-\sqrt{105}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{105}}{2*16}=\frac{19+\sqrt{105}}{32} $
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