4r^2+r-2=0

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Solution for 4r^2+r-2=0 equation:



4r^2+r-2=0
a = 4; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·4·(-2)
Δ = 33
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{-(1)-\sqrt{33}}{2*4}=\frac{-1-\sqrt{33}}{8} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{33}}{2*4}=\frac{-1+\sqrt{33}}{8} $

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