Solve Using the Quadratic Formula 2x(x-3)=15

$2 x (x - 3) = 15$

Step 1

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

Simplify the left side.

Simplify $2 x (x - 3)$ .

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 3 = 15$

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

Move $x$ .

$2 (x \cdot x) + 2 x \cdot - 3 = 15$

Multiply $x$ by $x$ .

$2 x^{2} + 2 x \cdot - 3 = 15$

Multiply $- 3$ by $2$ .

$2 x^{2} - 6 x = 15$

Subtract $15$ from both sides of the equation.

$2 x^{2} - 6 x - 15 = 0$

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 $a = 2$ , $b = - 6$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (2 \cdot - 15)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 2 \cdot - 15}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 15$ .

Multiply $- 4$ by $2$ .

$x = \frac{6 \pm \sqrt{36 - 8 \cdot - 15}}{2 \cdot 2}$

Multiply $- 8$ by $- 15$ .

$x = \frac{6 \pm \sqrt{36 + 120}}{2 \cdot 2}$

Add $36$ and $120$ .

$x = \frac{6 \pm \sqrt{156}}{2 \cdot 2}$

Rewrite $156$ as $2^{2} \cdot 39$ .

Factor $4$ out of $156$ .

$x = \frac{6 \pm \sqrt{4 (39)}}{2 \cdot 2}$

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 39}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{39}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{6 \pm 2 \sqrt{39}}{4}$

Simplify $\frac{6 \pm 2 \sqrt{39}}{4}$ .

$x = \frac{3 \pm \sqrt{39}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{3 \pm \sqrt{39}}{2}$

Decimal Form:

$x = 4.62249899 \dots, - 1.62249899 \dots$

Solve Using the Quadratic Formula 2x(x-3)=15

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