r^2+6r+7=4

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Solution for r^2+6r+7=4 equation:



r^2+6r+7=4
We move all terms to the left:
r^2+6r+7-(4)=0
We add all the numbers together, and all the variables
r^2+6r+3=0
a = 1; b = 6; c = +3;
Δ = b2-4ac
Δ = 62-4·1·3
Δ = 24
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:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{6}}{2*1}=\frac{-6-2\sqrt{6}}{2} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{6}}{2*1}=\frac{-6+2\sqrt{6}}{2} $

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