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3p^2-30p+48=0
a = 3; b = -30; c = +48;
Δ = b2-4ac
Δ = -302-4·3·48
Δ = 324
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{324}=18$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-18}{2*3}=\frac{12}{6} =2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+18}{2*3}=\frac{48}{6} =8 $
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