Solve for c (c)(4c+15)+4=0

$(c) (4 c + 15) + 4 = 0$

Step 1

Simplify each term.

Apply the distributive property.

$c (4 c) + c \cdot 15 + 4 = 0$

Rewrite using the commutative property of multiplication.

$4 c \cdot c + c \cdot 15 + 4 = 0$

Move $15$ to the left of $c$ .

$4 c \cdot c + 15 \cdot c + 4 = 0$

Multiply $c$ by $c$ by adding the exponents.

Move $c$ .

$4 (c \cdot c) + 15 \cdot c + 4 = 0$

Multiply $c$ by $c$ .

$4 c^{2} + 15 \cdot c + 4 = 0$

$4 c^{2} + 15 c + 4 = 0$

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 $a = 4$ , $b = 15$ , and $c = 4$ into the quadratic formula and solve for $c$ .

$\frac{- 15 \pm \sqrt{15^{2} - 4 \cdot (4 \cdot 4)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $15$ to the power of $2$ .

$c = \frac{- 15 \pm \sqrt{225 - 4 \cdot 4 \cdot 4}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$c = \frac{- 15 \pm \sqrt{225 - 16 \cdot 4}}{2 \cdot 4}$

Multiply $- 16$ by $4$ .

$c = \frac{- 15 \pm \sqrt{225 - 64}}{2 \cdot 4}$

Subtract $64$ from $225$ .

$c = \frac{- 15 \pm \sqrt{161}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$c = \frac{- 15 \pm \sqrt{161}}{8}$

Step 5

The final answer is the combination of both solutions.

$c = - \frac{15 - \sqrt{161}}{8}, - \frac{15 + \sqrt{161}}{8}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$c = - \frac{15 - \sqrt{161}}{8}, - \frac{15 + \sqrt{161}}{8}$

Decimal Form:

$c = - 0.28892780 \dots, - 3.46107219 \dots$