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36x^2-48x-19=0
a = 36; b = -48; c = -19;
Δ = b2-4ac
Δ = -482-4·36·(-19)
Δ = 5040
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5040}=\sqrt{144*35}=\sqrt{144}*\sqrt{35}=12\sqrt{35}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-12\sqrt{35}}{2*36}=\frac{48-12\sqrt{35}}{72} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+12\sqrt{35}}{2*36}=\frac{48+12\sqrt{35}}{72} $
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