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12p^2+9p-20=0
a = 12; b = 9; c = -20;
Δ = b2-4ac
Δ = 92-4·12·(-20)
Δ = 1041
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{1041}}{2*12}=\frac{-9-\sqrt{1041}}{24} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{1041}}{2*12}=\frac{-9+\sqrt{1041}}{24} $
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