Solve Using the Quadratic Formula 3x^2+x-1=(2x+9)(x-1)

$3 x^{2} + x - 1 = (2 x + 9) (x - 1)$

Step 1

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

Simplify the right side.

Simplify $(2 x + 9) (x - 1)$ .

Expand $(2 x + 9) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$3 x^{2} + x - 1 = 2 x (x - 1) + 9 (x - 1)$

Apply the distributive property.

$3 x^{2} + x - 1 = 2 x \cdot x + 2 x \cdot - 1 + 9 (x - 1)$

Apply the distributive property.

$3 x^{2} + x - 1 = 2 x \cdot x + 2 x \cdot - 1 + 9 x + 9 \cdot - 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$3 x^{2} + x - 1 = 2 (x \cdot x) + 2 x \cdot - 1 + 9 x + 9 \cdot - 1$

Multiply $x$ by $x$ .

$3 x^{2} + x - 1 = 2 x^{2} + 2 x \cdot - 1 + 9 x + 9 \cdot - 1$

Multiply $- 1$ by $2$ .

$3 x^{2} + x - 1 = 2 x^{2} - 2 x + 9 x + 9 \cdot - 1$

Multiply $9$ by $- 1$ .

$3 x^{2} + x - 1 = 2 x^{2} - 2 x + 9 x - 9$

Add $- 2 x$ and $9 x$ .

$3 x^{2} + x - 1 = 2 x^{2} + 7 x - 9$

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

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

$3 x^{2} + x - 1 - 2 x^{2} = 7 x - 9$

Subtract $7 x$ from both sides of the equation.

$3 x^{2} + x - 1 - 2 x^{2} - 7 x = - 9$

Add $9$ to both sides of the equation.

$3 x^{2} + x - 1 - 2 x^{2} - 7 x + 9 = 0$

Simplify $3 x^{2} + x - 1 - 2 x^{2} - 7 x + 9$ .

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

$x^{2} + x - 1 - 7 x + 9 = 0$

Subtract $7 x$ from $x$ .

$x^{2} - 6 x - 1 + 9 = 0$

Add $- 1$ and $9$ .

$x^{2} - 6 x + 8 = 0$

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 $a = 1$ , $b = - 6$ , and $c = 8$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 8)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 8$ .

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 8}}{2 \cdot 1}$

Multiply $- 4$ by $8$ .

$x = \frac{6 \pm \sqrt{36 - 32}}{2 \cdot 1}$

Subtract $32$ from $36$ .

$x = \frac{6 \pm \sqrt{4}}{2 \cdot 1}$

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

$x = \frac{6 \pm \sqrt{2^{2}}}{2 \cdot 1}$

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

$x = \frac{6 \pm 2}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{6 \pm 2}{2}$

Simplify $\frac{6 \pm 2}{2}$ .

$x = 3 \pm 1$

Step 5

The final answer is the combination of both solutions.

$x = 4, 2$

Solve Using the Quadratic Formula 3x^2+x-1=(2x+9)(x-1)

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