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(-13/65)q+100=0
Domain of the equation: 65)q!=0We multiply parentheses
q!=0/1
q!=0
q∈R
-13q^2+100=0
a = -13; b = 0; c = +100;
Δ = b2-4ac
Δ = 02-4·(-13)·100
Δ = 5200
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}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5200}=\sqrt{400*13}=\sqrt{400}*\sqrt{13}=20\sqrt{13}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{13}}{2*-13}=\frac{0-20\sqrt{13}}{-26} =-\frac{20\sqrt{13}}{-26} =-\frac{10\sqrt{13}}{-13} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{13}}{2*-13}=\frac{0+20\sqrt{13}}{-26} =\frac{20\sqrt{13}}{-26} =\frac{10\sqrt{13}}{-13} $
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