Solve 2x=sqrt{28x+29}-3

$$2x = \sqrt{ 28x+29 } -3$$

Solve for x

$$x=5$$

$$x=-1$$

Solution Steps

Subtract -3 from both sides of the equation.

$$2x+3=\sqrt{28x+29}$$

Square both sides of the equation.

$$\left(2x+3\right)^{2}=\left(\sqrt{28x+29}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$4x^{2}+12x+9=\left(\sqrt{28x+29}\right)^{2}$$

Calculate \sqrt{28x+29} to the power of 2 and get 28x+29.

$$4x^{2}+12x+9=28x+29$$

Subtract 28x from both sides.

$$4x^{2}+12x+9-28x=29$$

Combine 12x and -28x to get -16x.

$$4x^{2}-16x+9=29$$

Subtract 29 from both sides.

$$4x^{2}-16x+9-29=0$$

Subtract 29 from 9 to get -20.

$$4x^{2}-16x-20=0$$

Divide both sides by 4.

$$x^{2}-4x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-5\right)=-5$$

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. The only such pair is the system solution.

$$a=-5 b=1$$

Rewrite x^{2}-4x-5 as \left(x^{2}-5x\right)+\left(x-5\right).

$$\left(x^{2}-5x\right)+\left(x-5\right)$$

Factor out x in x^{2}-5x.

$$x\left(x-5\right)+x-5$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+1\right)$$

To find equation solutions, solve x-5=0 and x+1=0.

$$x=5 x=-1$$

Substitute 5 for x in the equation 2x=\sqrt{28x+29}-3.

$$2\times 5=\sqrt{28\times 5+29}-3$$

Simplify. The value x=5 satisfies the equation.

$$10=10$$

Substitute -1 for x in the equation 2x=\sqrt{28x+29}-3.

$$2\left(-1\right)=\sqrt{28\left(-1\right)+29}-3$$

Simplify. The value x=-1 satisfies the equation.

$$-2=-2$$

List all solutions of 2x+3=\sqrt{28x+29}.

$$x=5 x=-1$$