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12x^2-25x+12=0
a = 12; b = -25; c = +12;
Δ = b2-4ac
Δ = -252-4·12·12
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-7}{2*12}=\frac{18}{24} =3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+7}{2*12}=\frac{32}{24} =1+1/3 $
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