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\left(5x\right)^{2}=\left(\sqrt{10+15x}\right)^{2}
Square both sides of the equation.
5^{2}x^{2}=\left(\sqrt{10+15x}\right)^{2}
Expand \left(5x\right)^{2}.
25x^{2}=\left(\sqrt{10+15x}\right)^{2}
Calculate 5 to the power of 2 and get 25.
25x^{2}=10+15x
Calculate \sqrt{10+15x} to the power of 2 and get 10+15x.
25x^{2}-10=15x
Subtract 10 from both sides.
25x^{2}-10-15x=0
Subtract 15x from both sides.
5x^{2}-2-3x=0
Divide both sides by 5.
5x^{2}-3x-2=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-3 ab=5\left(-2\right)=-10
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-2. To find a and b, set up a system to be solved.
1,-10 2,-5
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.
1-10=-9 2-5=-3
Calculate the sum for each pair.
a=-5 b=2
The solution is the pair that gives sum -3.
\left(5x^{2}-5x\right)+\left(2x-2\right)
Rewrite 5x^{2}-3x-2 as \left(5x^{2}-5x\right)+\left(2x-2\right).
5x\left(x-1\right)+2\left(x-1\right)
Factor out 5x in the first and 2 in the second group.
\left(x-1\right)\left(5x+2\right)
Factor out common term x-1 by using distributive property.
x=1 x=-\frac{2}{5}
To find equation solutions, solve x-1=0 and 5x+2=0.
5\times 1=\sqrt{10+15\times 1}
Substitute 1 for x in the equation 5x=\sqrt{10+15x}.
5=5
Simplify. The value x=1 satisfies the equation.
5\left(-\frac{2}{5}\right)=\sqrt{10+15\left(-\frac{2}{5}\right)}
Substitute -\frac{2}{5} for x in the equation 5x=\sqrt{10+15x}.
-2=2
Simplify. The value x=-\frac{2}{5} does not satisfy the equation because the left and the right hand side have opposite signs.
x=1
Equation 5x=\sqrt{15x+10} has a unique solution.