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(8x^2)-25x+6=0
a = 8; b = -25; c = +6;
Δ = b2-4ac
Δ = -252-4·8·6
Δ = 433
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{-(-25)-\sqrt{433}}{2*8}=\frac{25-\sqrt{433}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{433}}{2*8}=\frac{25+\sqrt{433}}{16} $
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