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7p^2-49=0
a = 7; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·7·(-49)
Δ = 1372
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1372}=\sqrt{196*7}=\sqrt{196}*\sqrt{7}=14\sqrt{7}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{7}}{2*7}=\frac{0-14\sqrt{7}}{14} =-\frac{14\sqrt{7}}{14} =-\sqrt{7} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{7}}{2*7}=\frac{0+14\sqrt{7}}{14} =\frac{14\sqrt{7}}{14} =\sqrt{7} $
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