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12x^2-48x+21=0
a = 12; b = -48; c = +21;
Δ = b2-4ac
Δ = -482-4·12·21
Δ = 1296
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{1296}=36$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-36}{2*12}=\frac{12}{24} =1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+36}{2*12}=\frac{84}{24} =3+1/2 $
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