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12x^2-40x+12=0
a = 12; b = -40; c = +12;
Δ = b2-4ac
Δ = -402-4·12·12
Δ = 1024
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{1024}=32$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-32}{2*12}=\frac{8}{24} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+32}{2*12}=\frac{72}{24} =3 $
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