(r-2)(r+6)=64

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Solution for (r-2)(r+6)=64 equation:



(r-2)(r+6)=64
We move all terms to the left:
(r-2)(r+6)-(64)=0
We multiply parentheses ..
(+r^2+6r-2r-12)-64=0
We get rid of parentheses
r^2+6r-2r-12-64=0
We add all the numbers together, and all the variables
r^2+4r-76=0
a = 1; b = 4; c = -76;
Δ = b2-4ac
Δ = 42-4·1·(-76)
Δ = 320
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:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{5}}{2*1}=\frac{-4-8\sqrt{5}}{2} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{5}}{2*1}=\frac{-4+8\sqrt{5}}{2} $

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