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10x^2-48x+36=0
a = 10; b = -48; c = +36;
Δ = b2-4ac
Δ = -482-4·10·36
Δ = 864
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-12\sqrt{6}}{2*10}=\frac{48-12\sqrt{6}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+12\sqrt{6}}{2*10}=\frac{48+12\sqrt{6}}{20} $
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