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5x^2-20x+12.5=0
a = 5; b = -20; c = +12.5;
Δ = b2-4ac
Δ = -202-4·5·12.5
Δ = 150
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{150}=\sqrt{25*6}=\sqrt{25}*\sqrt{6}=5\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-5\sqrt{6}}{2*5}=\frac{20-5\sqrt{6}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+5\sqrt{6}}{2*5}=\frac{20+5\sqrt{6}}{10} $
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