Finite Math Examples

$y^{2} - 10 y = - 23$

Step 1

Add $23$ to both sides of the equation.

$y^{2} - 10 y + 23 = 0$

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 $a = 1$ , $b = - 10$ , and $c = 23$ into the quadratic formula and solve for $y$ .

$\frac{10 \pm \sqrt{{(- 10)}^{2} - 4 \cdot (1 \cdot 23)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 10$ to the power of $2$ .

$y = \frac{10 \pm \sqrt{100 - 4 \cdot 1 \cdot 23}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot 23$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$y = \frac{10 \pm \sqrt{100 - 4 \cdot 23}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $23$ .

$y = \frac{10 \pm \sqrt{100 - 92}}{2 \cdot 1}$

Step 4.1.3

Subtract $92$ from $100$ .

$y = \frac{10 \pm \sqrt{8}}{2 \cdot 1}$

Step 4.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 4.1.4.1

Factor $4$ out of $8$ .

$y = \frac{10 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Step 4.1.4.2

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

$y = \frac{10 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Step 4.1.5

Pull terms out from under the radical.

$y = \frac{10 \pm 2 \sqrt{2}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$y = \frac{10 \pm 2 \sqrt{2}}{2}$

Step 4.3

Simplify $\frac{10 \pm 2 \sqrt{2}}{2}$ .

$y = 5 \pm \sqrt{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$y = 5 \pm \sqrt{2}$

Decimal Form:

$y = 6.41421356 \dots, 3.58578643 \dots$