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19x^2-36=0
a = 19; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·19·(-36)
Δ = 2736
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2736}=\sqrt{144*19}=\sqrt{144}*\sqrt{19}=12\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{19}}{2*19}=\frac{0-12\sqrt{19}}{38} =-\frac{12\sqrt{19}}{38} =-\frac{6\sqrt{19}}{19} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{19}}{2*19}=\frac{0+12\sqrt{19}}{38} =\frac{12\sqrt{19}}{38} =\frac{6\sqrt{19}}{19} $
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