Solve for x 2x^2-5=3x^2-12x+31

$2 x^{2} - 5 = 3 x^{2} - 12 x + 31$

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$3 x^{2} - 12 x + 31 = 2 x^{2} - 5$

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

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

$3 x^{2} - 12 x + 31 - 2 x^{2} = - 5$

Subtract $2 x^{2}$ from $3 x^{2}$ .

$x^{2} - 12 x + 31 = - 5$

Add $5$ to both sides of the equation.

$x^{2} - 12 x + 31 + 5 = 0$

Add $31$ and $5$ .

$x^{2} - 12 x + 36 = 0$

Factor using the perfect square rule.

Rewrite $36$ as $6^{2}$ .

$x^{2} - 12 x + 6^{2} = 0$

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

$12 x = 2 \cdot x \cdot 6$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 6 + 6^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 6$ .

${(x - 6)}^{2} = 0$

Set the $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$