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q^2-13=80
We move all terms to the left:
q^2-13-(80)=0
We add all the numbers together, and all the variables
q^2-93=0
a = 1; b = 0; c = -93;
Δ = b2-4ac
Δ = 02-4·1·(-93)
Δ = 372
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}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{93}}{2*1}=\frac{0-2\sqrt{93}}{2} =-\frac{2\sqrt{93}}{2} =-\sqrt{93} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{93}}{2*1}=\frac{0+2\sqrt{93}}{2} =\frac{2\sqrt{93}}{2} =\sqrt{93} $
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