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\left(\sqrt{x+3}\right)^{2}=\left(3x-5\right)^{2}
Square both sides of the equation.
x+3=\left(3x-5\right)^{2}
Calculate \sqrt{x+3} to the power of 2 and get x+3.
x+3=9x^{2}-30x+25
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-5\right)^{2}.
x+3-9x^{2}=-30x+25
Subtract 9x^{2} from both sides.
x+3-9x^{2}+30x=25
Add 30x to both sides.
31x+3-9x^{2}=25
Combine x and 30x to get 31x.
31x+3-9x^{2}-25=0
Subtract 25 from both sides.
31x-22-9x^{2}=0
Subtract 25 from 3 to get -22.
-9x^{2}+31x-22=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=31 ab=-9\left(-22\right)=198
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx-22. To find a and b, set up a system to be solved.
1,198 2,99 3,66 6,33 9,22 11,18
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 198.
1+198=199 2+99=101 3+66=69 6+33=39 9+22=31 11+18=29
Calculate the sum for each pair.
a=22 b=9
The solution is the pair that gives sum 31.
\left(-9x^{2}+22x\right)+\left(9x-22\right)
Rewrite -9x^{2}+31x-22 as \left(-9x^{2}+22x\right)+\left(9x-22\right).
-x\left(9x-22\right)+9x-22
Factor out -x in -9x^{2}+22x.
\left(9x-22\right)\left(-x+1\right)
Factor out common term 9x-22 by using distributive property.
x=\frac{22}{9} x=1
To find equation solutions, solve 9x-22=0 and -x+1=0.
\sqrt{\frac{22}{9}+3}=3\times \frac{22}{9}-5
Substitute \frac{22}{9} for x in the equation \sqrt{x+3}=3x-5.
\frac{7}{3}=\frac{7}{3}
Simplify. The value x=\frac{22}{9} satisfies the equation.
\sqrt{1+3}=3\times 1-5
Substitute 1 for x in the equation \sqrt{x+3}=3x-5.
2=-2
Simplify. The value x=1 does not satisfy the equation because the left and the right hand side have opposite signs.
x=\frac{22}{9}
Equation \sqrt{x+3}=3x-5 has a unique solution.