4r^2+13r+5=0

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Solution for 4r^2+13r+5=0 equation: 



4r^2+13r+5=0

a = 4; b = 13; c = +5;
Δ = b2-4ac
Δ = 132-4·4·5
Δ = 89
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{89}}{2*4}=\frac{-13-\sqrt{89}}{8}
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{89}}{2*4}=\frac{-13+\sqrt{89}}{8}
$

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