-5r+r^2-14=0

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Solution for -5r+r^2-14=0 equation: 



-5r+r^2-14=0

a = 1; b = -5; c = -14;
Δ = b2-4ac
Δ = -52-4·1·(-14)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-9}{2*1}=\frac{-4}{2}
=-2
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+9}{2*1}=\frac{14}{2}
=7
$

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