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x^2-5x+5/4=0
We multiply all the terms by the denominator
x^2*4-5x*4+5=0
Wy multiply elements
4x^2-20x+5=0
a = 4; b = -20; c = +5;
Δ = b2-4ac
Δ = -202-4·4·5
Δ = 320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{5}}{2*4}=\frac{20-8\sqrt{5}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{5}}{2*4}=\frac{20+8\sqrt{5}}{8} $
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