Solve for x 5x^2-2x+16=4x^2+6x

$5 x^{2} - 2 x + 16 = 4 x^{2} + 6 x$

Move all terms containing $x$ to the left side of the equation.

Subtract $4 x^{2}$ from both sides of the equation.

$5 x^{2} - 2 x + 16 - 4 x^{2} = 6 x$

Subtract $6 x$ from both sides of the equation.

$5 x^{2} - 2 x + 16 - 4 x^{2} - 6 x = 0$

Subtract $4 x^{2}$ from $5 x^{2}$ .

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

Subtract $6 x$ from $- 2 x$ .

$x^{2} - 8 x + 16 = 0$

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$x^{2} - 8 x + 4^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 4 + 4^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 4$ .

${(x - 4)}^{2} = 0$

Set the $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$