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2x^2-20x+47=0
a = 2; b = -20; c = +47;
Δ = b2-4ac
Δ = -202-4·2·47
Δ = 24
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{6}}{2*2}=\frac{20-2\sqrt{6}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{6}}{2*2}=\frac{20+2\sqrt{6}}{4} $
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