r2+4r+1=0

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Solution for r2+4r+1=0 equation:



r2+4r+1=0
We add all the numbers together, and all the variables
r^2+4r+1=0
a = 1; b = 4; c = +1;
Δ = b2-4ac
Δ = 42-4·1·1
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{3}}{2*1}=\frac{-4-2\sqrt{3}}{2} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{3}}{2*1}=\frac{-4+2\sqrt{3}}{2} $

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