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16x^2-23x+8=0
a = 16; b = -23; c = +8;
Δ = b2-4ac
Δ = -232-4·16·8
Δ = 17
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{-(-23)-\sqrt{17}}{2*16}=\frac{23-\sqrt{17}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{17}}{2*16}=\frac{23+\sqrt{17}}{32} $
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