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p2-14p+48=0
We add all the numbers together, and all the variables
p^2-14p+48=0
a = 1; b = -14; c = +48;
Δ = b2-4ac
Δ = -142-4·1·48
Δ = 4
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{4}=2$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2}{2*1}=\frac{12}{2} =6 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2}{2*1}=\frac{16}{2} =8 $
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