Solve Using the Quadratic Formula t^2-3t=28

$t^{2} - 3 t = 28$

Step 1

Subtract $28$ from both sides of the equation.

$t^{2} - 3 t - 28 = 0$

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 $a = 1$ , $b = - 3$ , and $c = - 28$ into the quadratic formula and solve for $t$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (1 \cdot - 28)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$t = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 28}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 28$ .

Multiply $- 4$ by $1$ .

$t = \frac{3 \pm \sqrt{9 - 4 \cdot - 28}}{2 \cdot 1}$

Multiply $- 4$ by $- 28$ .

$t = \frac{3 \pm \sqrt{9 + 112}}{2 \cdot 1}$

Add $9$ and $112$ .

$t = \frac{3 \pm \sqrt{121}}{2 \cdot 1}$

Rewrite $121$ as $11^{2}$ .

$t = \frac{3 \pm \sqrt{11^{2}}}{2 \cdot 1}$

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

$t = \frac{3 \pm 11}{2 \cdot 1}$

Multiply $2$ by $1$ .

$t = \frac{3 \pm 11}{2}$

Step 5

The final answer is the combination of both solutions.

$t = 7, - 4$