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12x^2-50x-3=0
a = 12; b = -50; c = -3;
Δ = b2-4ac
Δ = -502-4·12·(-3)
Δ = 2644
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2644}=\sqrt{4*661}=\sqrt{4}*\sqrt{661}=2\sqrt{661}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{661}}{2*12}=\frac{50-2\sqrt{661}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{661}}{2*12}=\frac{50+2\sqrt{661}}{24} $
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