Solve Using the Quadratic Formula 2y^2-3y=1

$2 y^{2} - 3 y = 1$

Step 1

Subtract $1$ from both sides of the equation.

$2 y^{2} - 3 y - 1 = 0$

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 $a = 2$ , $b = - 3$ , and $c = - 1$ into the quadratic formula and solve for $y$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $2$ .

$y = \frac{3 \pm \sqrt{9 - 8 \cdot - 1}}{2 \cdot 2}$

Multiply $- 8$ by $- 1$ .

$y = \frac{3 \pm \sqrt{9 + 8}}{2 \cdot 2}$

Add $9$ and $8$ .

$y = \frac{3 \pm \sqrt{17}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$y = \frac{3 \pm \sqrt{17}}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$y = \frac{3 \pm \sqrt{17}}{4}$

Decimal Form:

$y = 1.78077640 \dots, - 0.28077640 \dots$