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\sqrt{7x+46}=x+4
Subtract -4 from both sides of the equation.
\left(\sqrt{7x+46}\right)^{2}=\left(x+4\right)^{2}
Square both sides of the equation.
7x+46=\left(x+4\right)^{2}
Calculate \sqrt{7x+46} to the power of 2 and get 7x+46.
7x+46=x^{2}+8x+16
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.
7x+46-x^{2}=8x+16
Subtract x^{2} from both sides.
7x+46-x^{2}-8x=16
Subtract 8x from both sides.
-x+46-x^{2}=16
Combine 7x and -8x to get -x.
-x+46-x^{2}-16=0
Subtract 16 from both sides.
-x+30-x^{2}=0
Subtract 16 from 46 to get 30.
-x^{2}-x+30=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-1 ab=-30=-30
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+30. To find a and b, set up a system to be solved.
1,-30 2,-15 3,-10 5,-6
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 -30.
1-30=-29 2-15=-13 3-10=-7 5-6=-1
Calculate the sum for each pair.
a=5 b=-6
The solution is the pair that gives sum -1.
\left(-x^{2}+5x\right)+\left(-6x+30\right)
Rewrite -x^{2}-x+30 as \left(-x^{2}+5x\right)+\left(-6x+30\right).
x\left(-x+5\right)+6\left(-x+5\right)
Factor out x in the first and 6 in the second group.
\left(-x+5\right)\left(x+6\right)
Factor out common term -x+5 by using distributive property.
x=5 x=-6
To find equation solutions, solve -x+5=0 and x+6=0.
\sqrt{7\times 5+46}-4=5
Substitute 5 for x in the equation \sqrt{7x+46}-4=x.
5=5
Simplify. The value x=5 satisfies the equation.
\sqrt{7\left(-6\right)+46}-4=-6
Substitute -6 for x in the equation \sqrt{7x+46}-4=x.
-2=-6
Simplify. The value x=-6 does not satisfy the equation.
x=5
Equation \sqrt{7x+46}=x+4 has a unique solution.