Solve Using the Quadratic Formula t^2+3=6t

$t^{2} + 3 = 6 t$

Step 1

Subtract $6 t$ from both sides of the equation.

$t^{2} + 3 - 6 t = 0$

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 $a = 1$ , $b = - 6$ , and $c = 3$ into the quadratic formula and solve for $t$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 3)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

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

$t = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 3$ .

Multiply $- 4$ by $1$ .

$t = \frac{6 \pm \sqrt{36 - 4 \cdot 3}}{2 \cdot 1}$

Multiply $- 4$ by $3$ .

$t = \frac{6 \pm \sqrt{36 - 12}}{2 \cdot 1}$

Subtract $12$ from $36$ .

$t = \frac{6 \pm \sqrt{24}}{2 \cdot 1}$

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

$t = \frac{6 \pm \sqrt{4 (6)}}{2 \cdot 1}$

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

$t = \frac{6 \pm \sqrt{2^{2} \cdot 6}}{2 \cdot 1}$

Pull terms out from under the radical.

$t = \frac{6 \pm 2 \sqrt{6}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$t = \frac{6 \pm 2 \sqrt{6}}{2}$

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

$t = 3 \pm \sqrt{6}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$t = 3 \pm \sqrt{6}$

Decimal Form:

$t = 5.44948974 \dots, 0.55051025 \dots$