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(12x^2)-13x+3=0
a = 12; b = -13; c = +3;
Δ = b2-4ac
Δ = -132-4·12·3
Δ = 25
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5}{2*12}=\frac{8}{24} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5}{2*12}=\frac{18}{24} =3/4 $
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