12r+r^2+20=0

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Solution for 12r+r^2+20=0 equation: 



12r+r^2+20=0

a = 1; b = 12; c = +20;
Δ = b2-4ac
Δ = 122-4·1·20
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*1}=\frac{-20}{2}
=-10
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*1}=\frac{-4}{2}
=-2
$

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