4r2-13r+6=0

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Solution for 4r2-13r+6=0 equation:



4r^2-13r+6=0
a = 4; b = -13; c = +6;
Δ = b2-4ac
Δ = -132-4·4·6
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{73}}{2*4}=\frac{13-\sqrt{73}}{8} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{73}}{2*4}=\frac{13+\sqrt{73}}{8} $

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