4x^2-20x+6=0

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Solution for 4x^2-20x+6=0 equation:



4x^2-20x+6=0
a = 4; b = -20; c = +6;
Δ = b2-4ac
Δ = -202-4·4·6
Δ = 304
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:
$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{19}}{2*4}=\frac{20-4\sqrt{19}}{8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{19}}{2*4}=\frac{20+4\sqrt{19}}{8} $

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