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12p^2-52p+56=0
a = 12; b = -52; c = +56;
Δ = b2-4ac
Δ = -522-4·12·56
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-4}{2*12}=\frac{48}{24} =2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+4}{2*12}=\frac{56}{24} =2+1/3 $
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