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15x^2-1050x+1000=0
a = 15; b = -1050; c = +1000;
Δ = b2-4ac
Δ = -10502-4·15·1000
Δ = 1042500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1042500}=\sqrt{2500*417}=\sqrt{2500}*\sqrt{417}=50\sqrt{417}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1050)-50\sqrt{417}}{2*15}=\frac{1050-50\sqrt{417}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1050)+50\sqrt{417}}{2*15}=\frac{1050+50\sqrt{417}}{30} $
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