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8x^2-20x+5=0
a = 8; b = -20; c = +5;
Δ = b2-4ac
Δ = -202-4·8·5
Δ = 240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{15}}{2*8}=\frac{20-4\sqrt{15}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{15}}{2*8}=\frac{20+4\sqrt{15}}{16} $
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