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12x^2-3x-14=0
a = 12; b = -3; c = -14;
Δ = b2-4ac
Δ = -32-4·12·(-14)
Δ = 681
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{681}}{2*12}=\frac{3-\sqrt{681}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{681}}{2*12}=\frac{3+\sqrt{681}}{24} $
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