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75=q^2
We move all terms to the left:
75-(q^2)=0
We add all the numbers together, and all the variables
-1q^2+75=0
a = -1; b = 0; c = +75;
Δ = b2-4ac
Δ = 02-4·(-1)·75
Δ = 300
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}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{3}}{2*-1}=\frac{0-10\sqrt{3}}{-2} =-\frac{10\sqrt{3}}{-2} =-\frac{5\sqrt{3}}{-1} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{3}}{2*-1}=\frac{0+10\sqrt{3}}{-2} =\frac{10\sqrt{3}}{-2} =\frac{5\sqrt{3}}{-1} $
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