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4p^2=60
We move all terms to the left:
4p^2-(60)=0
a = 4; b = 0; c = -60;
Δ = b2-4ac
Δ = 02-4·4·(-60)
Δ = 960
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{960}=\sqrt{64*15}=\sqrt{64}*\sqrt{15}=8\sqrt{15}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{15}}{2*4}=\frac{0-8\sqrt{15}}{8} =-\frac{8\sqrt{15}}{8} =-\sqrt{15} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{15}}{2*4}=\frac{0+8\sqrt{15}}{8} =\frac{8\sqrt{15}}{8} =\sqrt{15} $
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