4r^2+r-8=0

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Solution for 4r^2+r-8=0 equation: 



4r^2+r-8=0

a = 4; b = 1; c = -8;
Δ = b2-4ac
Δ = 12-4·4·(-8)
Δ = 129
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{129}}{2*4}=\frac{-1-\sqrt{129}}{8}
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{129}}{2*4}=\frac{-1+\sqrt{129}}{8}
$

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