Solve Using the Quadratic Formula 9(5-3x)^2-40=9

$9 {(5 - 3 x)}^{2} - 40 = 9$

Step 1

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

Subtract $9$ from both sides of the equation.

$9 {(5 - 3 x)}^{2} - 40 - 9 = 0$

Simplify $9 {(5 - 3 x)}^{2} - 40 - 9$ .

Simplify each term.

Rewrite ${(5 - 3 x)}^{2}$ as $(5 - 3 x) (5 - 3 x)$ .

$9 ((5 - 3 x) (5 - 3 x)) - 40 - 9 = 0$

Expand $(5 - 3 x) (5 - 3 x)$ using the FOIL Method.

Apply the distributive property.

$9 (5 (5 - 3 x) - 3 x (5 - 3 x)) - 40 - 9 = 0$

Apply the distributive property.

$9 (5 \cdot 5 + 5 (- 3 x) - 3 x (5 - 3 x)) - 40 - 9 = 0$

Apply the distributive property.

$9 (5 \cdot 5 + 5 (- 3 x) - 3 x \cdot 5 - 3 x (- 3 x)) - 40 - 9 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $5$ by $5$ .

$9 (25 + 5 (- 3 x) - 3 x \cdot 5 - 3 x (- 3 x)) - 40 - 9 = 0$

Multiply $- 3$ by $5$ .

$9 (25 - 15 x - 3 x \cdot 5 - 3 x (- 3 x)) - 40 - 9 = 0$

Multiply $5$ by $- 3$ .

$9 (25 - 15 x - 15 x - 3 x (- 3 x)) - 40 - 9 = 0$

Rewrite using the commutative property of multiplication.

$9 (25 - 15 x - 15 x - 3 \cdot - 3 x \cdot x) - 40 - 9 = 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$9 (25 - 15 x - 15 x - 3 \cdot - 3 (x \cdot x)) - 40 - 9 = 0$

Multiply $x$ by $x$ .

$9 (25 - 15 x - 15 x - 3 \cdot - 3 x^{2}) - 40 - 9 = 0$

Multiply $- 3$ by $- 3$ .

$9 (25 - 15 x - 15 x + 9 x^{2}) - 40 - 9 = 0$

Subtract $15 x$ from $- 15 x$ .

$9 (25 - 30 x + 9 x^{2}) - 40 - 9 = 0$

Apply the distributive property.

$9 \cdot 25 + 9 (- 30 x) + 9 (9 x^{2}) - 40 - 9 = 0$

Simplify.

Multiply $9$ by $25$ .

$225 + 9 (- 30 x) + 9 (9 x^{2}) - 40 - 9 = 0$

Multiply $- 30$ by $9$ .

$225 - 270 x + 9 (9 x^{2}) - 40 - 9 = 0$

Multiply $9$ by $9$ .

$225 - 270 x + 81 x^{2} - 40 - 9 = 0$

Subtract $40$ from $225$ .

$- 270 x + 81 x^{2} + 185 - 9 = 0$

Subtract $9$ from $185$ .

$- 270 x + 81 x^{2} + 176 = 0$

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 $a = 81$ , $b = - 270$ , and $c = 176$ into the quadratic formula and solve for $x$ .

$\frac{270 \pm \sqrt{{(- 270)}^{2} - 4 \cdot (81 \cdot 176)}}{2 \cdot 81}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 270$ to the power of $2$ .

$x = \frac{270 \pm \sqrt{72900 - 4 \cdot 81 \cdot 176}}{2 \cdot 81}$

Multiply $- 4 \cdot 81 \cdot 176$ .

Multiply $- 4$ by $81$ .

$x = \frac{270 \pm \sqrt{72900 - 324 \cdot 176}}{2 \cdot 81}$

Multiply $- 324$ by $176$ .

$x = \frac{270 \pm \sqrt{72900 - 57024}}{2 \cdot 81}$

Subtract $57024$ from $72900$ .

$x = \frac{270 \pm \sqrt{15876}}{2 \cdot 81}$

Rewrite $15876$ as $126^{2}$ .

$x = \frac{270 \pm \sqrt{126^{2}}}{2 \cdot 81}$

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

$x = \frac{270 \pm 126}{2 \cdot 81}$

Multiply $2$ by $81$ .

$x = \frac{270 \pm 126}{162}$

Simplify $\frac{270 \pm 126}{162}$ .

$x = \frac{15 \pm 7}{9}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{22}{9}, \frac{8}{9}$

Solve Using the Quadratic Formula 9(5-3x)^2-40=9

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