q2+2=19

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Solution for q2+2=19 equation:



q2+2=19
We move all terms to the left:
q2+2-(19)=0
We add all the numbers together, and all the variables
q^2-17=0
a = 1; b = 0; c = -17;
Δ = b2-4ac
Δ = 02-4·1·(-17)
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{17}}{2*1}=\frac{0-2\sqrt{17}}{2} =-\frac{2\sqrt{17}}{2} =-\sqrt{17} $
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{17}}{2*1}=\frac{0+2\sqrt{17}}{2} =\frac{2\sqrt{17}}{2} =\sqrt{17} $

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