Solve Using the Quadratic Formula y^2+5y-4=0

$y^{2} + 5 y - 4 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 1$ , $b = 5$ , and $c = - 4$ into the quadratic formula and solve for $y$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot - 4)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $5$ to the power of $2$ .

$y = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$y = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 4}}{2 \cdot 1}$

Multiply $- 4$ by $- 4$ .

$y = \frac{- 5 \pm \sqrt{25 + 16}}{2 \cdot 1}$

Add $25$ and $16$ .

$y = \frac{- 5 \pm \sqrt{41}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{- 5 \pm \sqrt{41}}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$y = \frac{- 5 \pm \sqrt{41}}{2}$

Decimal Form:

$y = 0.70156211 \dots, - 5.70156211 \dots$