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7p^2+p-10=0
a = 7; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·7·(-10)
Δ = 281
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{-(1)-\sqrt{281}}{2*7}=\frac{-1-\sqrt{281}}{14} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{281}}{2*7}=\frac{-1+\sqrt{281}}{14} $
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