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12x^2-50x+25=0
a = 12; b = -50; c = +25;
Δ = b2-4ac
Δ = -502-4·12·25
Δ = 1300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-10\sqrt{13}}{2*12}=\frac{50-10\sqrt{13}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+10\sqrt{13}}{2*12}=\frac{50+10\sqrt{13}}{24} $
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