Find All Complex Solutions 2x^2-5x=x^2-6x+2

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

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.

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

Add $6 x$ to both sides of the equation.

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

Subtract $2$ from both sides of the equation.

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

Simplify $2 x^{2} - 5 x - x^{2} + 6 x - 2$ .

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

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

Add $- 5 x$ and $6 x$ .

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

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 $a = 1$ , $b = 1$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 2)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 2$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4$ by $- 2$ .

$x = \frac{- 1 \pm \sqrt{1 + 8}}{2 \cdot 1}$

Add $1$ and $8$ .

$x = \frac{- 1 \pm \sqrt{9}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{- 1 \pm \sqrt{3^{2}}}{2 \cdot 1}$

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

$x = \frac{- 1 \pm 3}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm 3}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 1, - 2$