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2p^2-5p-42=0
a = 2; b = -5; c = -42;
Δ = b2-4ac
Δ = -52-4·2·(-42)
Δ = 361
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{361}=19$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-19}{2*2}=\frac{-14}{4} =-3+1/2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+19}{2*2}=\frac{24}{4} =6 $
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