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\left(2p\right)^{2}=\left(\sqrt{3p^{2}+\left(5-\frac{3}{2}p\right)^{2}}\right)^{2}
Square both sides of the equation.
2^{2}p^{2}=\left(\sqrt{3p^{2}+\left(5-\frac{3}{2}p\right)^{2}}\right)^{2}
Expand \left(2p\right)^{2}.
4p^{2}=\left(\sqrt{3p^{2}+\left(5-\frac{3}{2}p\right)^{2}}\right)^{2}
Calculate 2 to the power of 2 and get 4.
4p^{2}=\left(\sqrt{3p^{2}+25-15p+\frac{9}{4}p^{2}}\right)^{2}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\frac{3}{2}p\right)^{2}.
4p^{2}=\left(\sqrt{\frac{21}{4}p^{2}+25-15p}\right)^{2}
Combine 3p^{2} and \frac{9}{4}p^{2} to get \frac{21}{4}p^{2}.
4p^{2}=\frac{21}{4}p^{2}+25-15p
Calculate \sqrt{\frac{21}{4}p^{2}+25-15p} to the power of 2 and get \frac{21}{4}p^{2}+25-15p.
4p^{2}-\frac{21}{4}p^{2}=25-15p
Subtract \frac{21}{4}p^{2} from both sides.
-\frac{5}{4}p^{2}=25-15p
Combine 4p^{2} and -\frac{21}{4}p^{2} to get -\frac{5}{4}p^{2}.
-\frac{5}{4}p^{2}-25=-15p
Subtract 25 from both sides.
-\frac{5}{4}p^{2}-25+15p=0
Add 15p to both sides.
-\frac{5}{4}p^{2}+15p-25=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
p=\frac{-15±\sqrt{15^{2}-4\left(-\frac{5}{4}\right)\left(-25\right)}}{2\left(-\frac{5}{4}\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{5}{4} for a, 15 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
p=\frac{-15±\sqrt{225-4\left(-\frac{5}{4}\right)\left(-25\right)}}{2\left(-\frac{5}{4}\right)}
Square 15.
p=\frac{-15±\sqrt{225+5\left(-25\right)}}{2\left(-\frac{5}{4}\right)}
Multiply -4 times -\frac{5}{4}.
p=\frac{-15±\sqrt{225-125}}{2\left(-\frac{5}{4}\right)}
Multiply 5 times -25.
p=\frac{-15±\sqrt{100}}{2\left(-\frac{5}{4}\right)}
Add 225 to -125.
p=\frac{-15±10}{2\left(-\frac{5}{4}\right)}
Take the square root of 100.
p=\frac{-15±10}{-\frac{5}{2}}
Multiply 2 times -\frac{5}{4}.
p=-\frac{5}{-\frac{5}{2}}
Now solve the equation p=\frac{-15±10}{-\frac{5}{2}} when ± is plus. Add -15 to 10.
p=2
Divide -5 by -\frac{5}{2} by multiplying -5 by the reciprocal of -\frac{5}{2}.
p=-\frac{25}{-\frac{5}{2}}
Now solve the equation p=\frac{-15±10}{-\frac{5}{2}} when ± is minus. Subtract 10 from -15.
p=10
Divide -25 by -\frac{5}{2} by multiplying -25 by the reciprocal of -\frac{5}{2}.
p=2 p=10
The equation is now solved.
2\times 2=\sqrt{3\times 2^{2}+\left(5-\frac{3}{2}\times 2\right)^{2}}
Substitute 2 for p in the equation 2p=\sqrt{3p^{2}+\left(5-\frac{3}{2}p\right)^{2}}.
4=4
Simplify. The value p=2 satisfies the equation.
2\times 10=\sqrt{3\times 10^{2}+\left(5-\frac{3}{2}\times 10\right)^{2}}
Substitute 10 for p in the equation 2p=\sqrt{3p^{2}+\left(5-\frac{3}{2}p\right)^{2}}.
20=20
Simplify. The value p=10 satisfies the equation.
p=2 p=10
List all solutions of 2p=\sqrt{\left(-\frac{3p}{2}+5\right)^{2}+3p^{2}}.