9r+r^2+8=0

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Solution for 9r+r^2+8=0 equation:



9r+r^2+8=0
a = 1; b = 9; c = +8;
Δ = b2-4ac
Δ = 92-4·1·8
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-7}{2*1}=\frac{-16}{2} =-8 $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+7}{2*1}=\frac{-2}{2} =-1 $

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