Precalculus Examples

$y^{2} + y - 2400 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 1$ , $b = 1$ , and $c = - 2400$ into the quadratic formula and solve for $y$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 2400)}}{2 \cdot 1}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

One to any power is one.

$y = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 2400}}{2 \cdot 1}$

Step 3.1.2

Multiply $- 4 \cdot 1 \cdot - 2400$ .

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Step 3.1.2.1

Multiply $- 4$ by $1$ .

$y = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 2400}}{2 \cdot 1}$

Step 3.1.2.2

Multiply $- 4$ by $- 2400$ .

$y = \frac{- 1 \pm \sqrt{1 + 9600}}{2 \cdot 1}$

Step 3.1.3

Add $1$ and $9600$ .

$y = \frac{- 1 \pm \sqrt{9601}}{2 \cdot 1}$

Step 3.2

Multiply $2$ by $1$ .

$y = \frac{- 1 \pm \sqrt{9601}}{2}$

Step 4

The final answer is the combination of both solutions.

$y = - \frac{1 - \sqrt{9601}}{2}, - \frac{1 + \sqrt{9601}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$y = - \frac{1 - \sqrt{9601}}{2}, - \frac{1 + \sqrt{9601}}{2}$

Decimal Form:

$y = 48.49234634 \dots, - 49.49234634 \dots$