Solve Using the Quadratic Formula 13x^2-16x=x^2-x

$13 x^{2} - 16 x = x^{2} - x$

Step 1

Move all terms to the left side of the equation and simplify.

Move all the expressions to the left side of the equation.

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

$13 x^{2} - 16 x - x^{2} = - x$

Add $x$ to both sides of the equation.

$13 x^{2} - 16 x - x^{2} + x = 0$

Simplify $13 x^{2} - 16 x - x^{2} + x$ .

Subtract $x^{2}$ from $13 x^{2}$ .

$12 x^{2} - 16 x + x = 0$

Add $- 16 x$ and $x$ .

$12 x^{2} - 15 x = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 12$ , $b = - 15$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{15 \pm \sqrt{{(- 15)}^{2} - 4 \cdot (12 \cdot 0)}}{2 \cdot 12}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 15$ to the power of $2$ .

$x = \frac{15 \pm \sqrt{225 - 4 \cdot 12 \cdot 0}}{2 \cdot 12}$

Multiply $- 4 \cdot 12 \cdot 0$ .

Multiply $- 4$ by $12$ .

$x = \frac{15 \pm \sqrt{225 - 48 \cdot 0}}{2 \cdot 12}$

Multiply $- 48$ by $0$ .

$x = \frac{15 \pm \sqrt{225 + 0}}{2 \cdot 12}$

Add $225$ and $0$ .

$x = \frac{15 \pm \sqrt{225}}{2 \cdot 12}$

Rewrite $225$ as $15^{2}$ .

$x = \frac{15 \pm \sqrt{15^{2}}}{2 \cdot 12}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{15 \pm 15}{2 \cdot 12}$

Multiply $2$ by $12$ .

$x = \frac{15 \pm 15}{24}$

Simplify $\frac{15 \pm 15}{24}$ .

$x = \frac{5 \pm 5}{8}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{5}{4}, 0$