Solve Using the Quadratic Formula (3-y)(y+4)=3y-5

$(3 - y) (y + 4) = 3 y - 5$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify $(3 - y) (y + 4)$ .

Expand $(3 - y) (y + 4)$ using the FOIL Method.

Apply the distributive property.

$3 (y + 4) - y (y + 4) = 3 y - 5$

Apply the distributive property.

$3 y + 3 \cdot 4 - y (y + 4) = 3 y - 5$

Apply the distributive property.

$3 y + 3 \cdot 4 - y \cdot y - y \cdot 4 = 3 y - 5$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $4$ .

$3 y + 12 - y \cdot y - y \cdot 4 = 3 y - 5$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$3 y + 12 - (y \cdot y) - y \cdot 4 = 3 y - 5$

Multiply $y$ by $y$ .

$3 y + 12 - y^{2} - y \cdot 4 = 3 y - 5$

Multiply $4$ by $- 1$ .

$3 y + 12 - y^{2} - 4 y = 3 y - 5$

Subtract $4 y$ from $3 y$ .

$- y + 12 - y^{2} = 3 y - 5$

Move all the expressions to the left side of the equation.

Subtract $3 y$ from both sides of the equation.

$- y + 12 - y^{2} - 3 y = - 5$

Add $5$ to both sides of the equation.

$- y + 12 - y^{2} - 3 y + 5 = 0$

Simplify $- y + 12 - y^{2} - 3 y + 5$ .

Subtract $3 y$ from $- y$ .

$- 4 y + 12 - y^{2} + 5 = 0$

Add $12$ and $5$ .

$- 4 y - y^{2} + 17 = 0$

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 $a = - 1$ , $b = - 4$ , and $c = 17$ into the quadratic formula and solve for $y$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (- 1 \cdot 17)}}{2 \cdot - 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$y = \frac{4 \pm \sqrt{16 - 4 \cdot - 1 \cdot 17}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot 17$ .

Multiply $- 4$ by $- 1$ .

$y = \frac{4 \pm \sqrt{16 + 4 \cdot 17}}{2 \cdot - 1}$

Multiply $4$ by $17$ .

$y = \frac{4 \pm \sqrt{16 + 68}}{2 \cdot - 1}$

Add $16$ and $68$ .

$y = \frac{4 \pm \sqrt{84}}{2 \cdot - 1}$

Rewrite $84$ as $2^{2} \cdot 21$ .

Factor $4$ out of $84$ .

$y = \frac{4 \pm \sqrt{4 (21)}}{2 \cdot - 1}$

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

$y = \frac{4 \pm \sqrt{2^{2} \cdot 21}}{2 \cdot - 1}$

Pull terms out from under the radical.

$y = \frac{4 \pm 2 \sqrt{21}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$y = \frac{4 \pm 2 \sqrt{21}}{- 2}$

Simplify $\frac{4 \pm 2 \sqrt{21}}{- 2}$ .

$y = \frac{2 \pm \sqrt{21}}{- 1}$

Move the negative one from the denominator of $\frac{2 \pm \sqrt{21}}{- 1}$ .

$y = - 1 \cdot (2 \pm \sqrt{21})$

Rewrite $- 1 \cdot (2 \pm \sqrt{21})$ as $- (2 \pm \sqrt{21})$ .

$y = - (2 \pm \sqrt{21})$

Step 5

The result can be shown in multiple forms.

Exact Form:

$y = - (2 \pm \sqrt{21})$

Decimal Form:

$y = - 6.58257569 \dots, 2.58257569 \dots$

Solve Using the Quadratic Formula (3-y)(y+4)=3y-5

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