Solve Using the Quadratic Formula 2t^2+1=6t

$2 t^{2} + 1 = 6 t$

Step 1

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

$2 t^{2} + 1 - 6 t = 0$

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 $a = 2$ , $b = - 6$ , and $c = 1$ into the quadratic formula and solve for $t$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 2 \cdot 1$ .

Multiply $- 4$ by $2$ .

$t = \frac{6 \pm \sqrt{36 - 8 \cdot 1}}{2 \cdot 2}$

Multiply $- 8$ by $1$ .

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

Subtract $8$ from $36$ .

$t = \frac{6 \pm \sqrt{28}}{2 \cdot 2}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$t = \frac{6 \pm \sqrt{4 (7)}}{2 \cdot 2}$

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

$t = \frac{6 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 2}$

Pull terms out from under the radical.

$t = \frac{6 \pm 2 \sqrt{7}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$t = \frac{6 \pm 2 \sqrt{7}}{4}$

Simplify $\frac{6 \pm 2 \sqrt{7}}{4}$ .

$t = \frac{3 \pm \sqrt{7}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$t = \frac{3 \pm \sqrt{7}}{2}$

Decimal Form:

$t = 2.82287565 \dots, 0.17712434 \dots$