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\sqrt{5x+19}=-1-x
Subtract x from both sides of the equation.
\left(\sqrt{5x+19}\right)^{2}=\left(-1-x\right)^{2}
Square both sides of the equation.
5x+19=\left(-1-x\right)^{2}
Calculate \sqrt{5x+19} to the power of 2 and get 5x+19.
5x+19=1+2x+x^{2}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-1-x\right)^{2}.
5x+19-1=2x+x^{2}
Subtract 1 from both sides.
5x+18=2x+x^{2}
Subtract 1 from 19 to get 18.
5x+18-2x=x^{2}
Subtract 2x from both sides.
3x+18=x^{2}
Combine 5x and -2x to get 3x.
3x+18-x^{2}=0
Subtract x^{2} from both sides.
-x^{2}+3x+18=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=3 ab=-18=-18
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+18. To find a and b, set up a system to be solved.
-1,18 -2,9 -3,6
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.
-1+18=17 -2+9=7 -3+6=3
Calculate the sum for each pair.
a=6 b=-3
The solution is the pair that gives sum 3.
\left(-x^{2}+6x\right)+\left(-3x+18\right)
Rewrite -x^{2}+3x+18 as \left(-x^{2}+6x\right)+\left(-3x+18\right).
-x\left(x-6\right)-3\left(x-6\right)
Factor out -x in the first and -3 in the second group.
\left(x-6\right)\left(-x-3\right)
Factor out common term x-6 by using distributive property.
x=6 x=-3
To find equation solutions, solve x-6=0 and -x-3=0.
6+\sqrt{5\times 6+19}=-1
Substitute 6 for x in the equation x+\sqrt{5x+19}=-1.
13=-1
Simplify. The value x=6 does not satisfy the equation because the left and the right hand side have opposite signs.
-3+\sqrt{5\left(-3\right)+19}=-1
Substitute -3 for x in the equation x+\sqrt{5x+19}=-1.
-1=-1
Simplify. The value x=-3 satisfies the equation.
x=-3
Equation \sqrt{5x+19}=-x-1 has a unique solution.