Solve Using the Quadratic Formula x(x+1)=306

$x (x + 1) = 306$

Step 1

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

Simplify the left side.

Simplify $x (x + 1)$ .

Apply the distributive property.

$x \cdot x + x \cdot 1 = 306$

Simplify the expression.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 1 = 306$

Multiply $x$ by $1$ .

$x^{2} + x = 306$

Subtract $306$ from both sides of the equation.

$x^{2} + x - 306 = 0$

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 $c = - 306$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 306)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 306}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 306$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 306}}{2 \cdot 1}$

Multiply $- 4$ by $- 306$ .

$x = \frac{- 1 \pm \sqrt{1 + 1224}}{2 \cdot 1}$

Add $1$ and $1224$ .

$x = \frac{- 1 \pm \sqrt{1225}}{2 \cdot 1}$

Rewrite $1225$ as $35^{2}$ .

$x = \frac{- 1 \pm \sqrt{35^{2}}}{2 \cdot 1}$

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

$x = \frac{- 1 \pm 35}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm 35}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 17, - 18$