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12x^2-280x+1125=0
a = 12; b = -280; c = +1125;
Δ = b2-4ac
Δ = -2802-4·12·1125
Δ = 24400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24400}=\sqrt{400*61}=\sqrt{400}*\sqrt{61}=20\sqrt{61}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-280)-20\sqrt{61}}{2*12}=\frac{280-20\sqrt{61}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-280)+20\sqrt{61}}{2*12}=\frac{280+20\sqrt{61}}{24} $
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