Solve each equation. (x+4)^2=3x+40

To solve the equation $(x+4)^{2}=3x+40 side of the equation:<br /><br />$(x+4)^{2} = x^2 + 8x + 16$<br /><br />So the equation becomes:<br /><br />$x^2 + 8x + 16 = 3x + 40$<br /><br />Next, we move all terms to one side of the equation to set it equal to zero:<br /><br />$x^2 + 8x + 16 - 3x - 40 = 0$<br /><br />Simplifying, we get:<br /><br />$x^2 + 5x - 24 = 0$<br /><br />Now, we can factor the quadratic equation:<br /><br />$(x + 8)(x - 3) = 0$<br /><br />Setting each factor equal to zero gives us the solutions:<br /><br />$x + 8 = 0$ or $x - 3 = 0$<br /><br />Solving for $x$, we find:<br /><br />$x = -8$ or $x = 3$<br /><br />Therefore, $(x+4)^{2}=3x+40$ are $x = -8$ and $x = 3$.