Solve the Square Root Equation: 3x^2+10x-16 = 2x^2+4x

To solve this problem, we'll follow these steps:

• Step 1: Eliminate the square roots by squaring both sides of the equation.
• Step 2: Simplify the resulting equation and bring all terms to one side.
• Step 3: Factor the resulting quadratic equation.
• Step 4: Solve for $x$ and check for valid solutions by substituting back into the original equation.

Now, let's work through each step in detail:

Step 1: Square both sides of the equation:

$\left(\sqrt{3x^2 + 10x - 16}\right)^2 = \left(\sqrt{2x^2 + 4x}\right)^2$

The equation becomes:

$3x^2 + 10x - 16 = 2x^2 + 4x$

Step 2: Simplify and rearrange the equation:

$3x^2 + 10x - 16 - 2x^2 - 4x = 0$

This simplifies to:

$x^2 + 6x - 16 = 0$

Step 3: Factor the quadratic equation:

We need to find two numbers that multiply to $-16$ and add to $6$. These numbers are $8$ and $-2$.

The equation factors to:

$(x + 8)(x - 2) = 0$

Step 4: Solve for $x$:

Set each factor equal to zero:

• $x + 8 = 0 \implies x = -8$
• $x - 2 = 0 \implies x = 2$

Finally, check these solutions in the original equation:

• For $x = -8$, $\sqrt{3(-8)^2 + 10(-8) - 16} = \sqrt{2(-8)^2 + 4(-8)}$

$\sqrt{192 - 80 - 16} = \sqrt{128 - 32}$

$\sqrt{96} = \sqrt{96}$

• For $x = 2$, $\sqrt{3(2)^2 + 10(2) - 16} = \sqrt{2(2)^2 + 4(2)}$

$\sqrt{12 + 20 - 16} = \sqrt{8 + 8}$

$\sqrt{16} = \sqrt{16}$

Both solutions are valid. Therefore, the solutions to the equation are:

$x = -8$ and $x = 2$.

The correct choice is $\textbf{d}: -8, 2$.