Solve Using the Quadratic Formula x(2x-5)=12

$x (2 x - 5) = 12$

Step 1

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

Simplify the left side.

Simplify $x (2 x - 5)$ .

Simplify by multiplying through.

Apply the distributive property.

$x (2 x) + x \cdot - 5 = 12$

Reorder.

Rewrite using the commutative property of multiplication.

$2 x \cdot x + x \cdot - 5 = 12$

Move $- 5$ to the left of $x$ .

$2 x \cdot x - 5 \cdot x = 12$

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

Move $x$ .

$2 (x \cdot x) - 5 \cdot x = 12$

Multiply $x$ by $x$ .

$2 x^{2} - 5 \cdot x = 12$

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

Subtract $12$ from both sides of the equation.

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

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 $a = 2$ , $b = - 5$ , and $c = - 12$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (2 \cdot - 12)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 2 \cdot - 12}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 12$ .

Multiply $- 4$ by $2$ .

$x = \frac{5 \pm \sqrt{25 - 8 \cdot - 12}}{2 \cdot 2}$

Multiply $- 8$ by $- 12$ .

$x = \frac{5 \pm \sqrt{25 + 96}}{2 \cdot 2}$

Add $25$ and $96$ .

$x = \frac{5 \pm \sqrt{121}}{2 \cdot 2}$

Rewrite $121$ as $11^{2}$ .

$x = \frac{5 \pm \sqrt{11^{2}}}{2 \cdot 2}$

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

$x = \frac{5 \pm 11}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{5 \pm 11}{4}$

Step 5

The final answer is the combination of both solutions.

$x = 4, - \frac{3}{2}$

Solve Using the Quadratic Formula x(2x-5)=12

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