Solve Using the Quadratic Formula 13-6x=(2x-5)^2+3

$13 - 6 x = {(2 x - 5)}^{2} + 3$

Step 1

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

Simplify the right side.

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

Simplify each term.

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

$13 - 6 x = (2 x - 5) (2 x - 5) + 3$

Expand $(2 x - 5) (2 x - 5)$ using the FOIL Method.

Apply the distributive property.

$13 - 6 x = 2 x (2 x - 5) - 5 (2 x - 5) + 3$

Apply the distributive property.

$13 - 6 x = 2 x (2 x) + 2 x \cdot - 5 - 5 (2 x - 5) + 3$

Apply the distributive property.

$13 - 6 x = 2 x (2 x) + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5 + 3$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$13 - 6 x = 2 \cdot 2 x \cdot x + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5 + 3$

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

Move $x$ .

$13 - 6 x = 2 \cdot 2 (x \cdot x) + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5 + 3$

Multiply $x$ by $x$ .

$13 - 6 x = 2 \cdot 2 x^{2} + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5 + 3$

Multiply $2$ by $2$ .

$13 - 6 x = 4 x^{2} + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5 + 3$

Multiply $- 5$ by $2$ .

$13 - 6 x = 4 x^{2} - 10 x - 5 (2 x) - 5 \cdot - 5 + 3$

Multiply $2$ by $- 5$ .

$13 - 6 x = 4 x^{2} - 10 x - 10 x - 5 \cdot - 5 + 3$

Multiply $- 5$ by $- 5$ .

$13 - 6 x = 4 x^{2} - 10 x - 10 x + 25 + 3$

Subtract $10 x$ from $- 10 x$ .

$13 - 6 x = 4 x^{2} - 20 x + 25 + 3$

Add $25$ and $3$ .

$13 - 6 x = 4 x^{2} - 20 x + 28$

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

Subtract $4 x^{2}$ from both sides of the equation.

$13 - 6 x - 4 x^{2} = - 20 x + 28$

Add $20 x$ to both sides of the equation.

$13 - 6 x - 4 x^{2} + 20 x = 28$

Subtract $28$ from both sides of the equation.

$13 - 6 x - 4 x^{2} + 20 x - 28 = 0$

Simplify $13 - 6 x - 4 x^{2} + 20 x - 28$ .

Subtract $28$ from $13$ .

$- 6 x - 4 x^{2} + 20 x - 15 = 0$

Add $- 6 x$ and $20 x$ .

$- 4 x^{2} + 14 x - 15 = 0$

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 $a = - 4$ , $b = 14$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot (- 4 \cdot - 15)}}{2 \cdot - 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $14$ to the power of $2$ .

$x = \frac{- 14 \pm \sqrt{196 - 4 \cdot - 4 \cdot - 15}}{2 \cdot - 4}$

Multiply $- 4 \cdot - 4 \cdot - 15$ .

Multiply $- 4$ by $- 4$ .

$x = \frac{- 14 \pm \sqrt{196 + 16 \cdot - 15}}{2 \cdot - 4}$

Multiply $16$ by $- 15$ .

$x = \frac{- 14 \pm \sqrt{196 - 240}}{2 \cdot - 4}$

Subtract $240$ from $196$ .

$x = \frac{- 14 \pm \sqrt{- 44}}{2 \cdot - 4}$

Rewrite $- 44$ as $- 1 (44)$ .

$x = \frac{- 14 \pm \sqrt{- 1 \cdot 44}}{2 \cdot - 4}$

Rewrite $\sqrt{- 1 (44)}$ as $\sqrt{- 1} \cdot \sqrt{44}$ .

$x = \frac{- 14 \pm \sqrt{- 1} \cdot \sqrt{44}}{2 \cdot - 4}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 14 \pm i \cdot \sqrt{44}}{2 \cdot - 4}$

Rewrite $44$ as $2^{2} \cdot 11$ .

Factor $4$ out of $44$ .

$x = \frac{- 14 \pm i \cdot \sqrt{4 (11)}}{2 \cdot - 4}$

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

$x = \frac{- 14 \pm i \cdot \sqrt{2^{2} \cdot 11}}{2 \cdot - 4}$

Pull terms out from under the radical.

$x = \frac{- 14 \pm i \cdot (2 \sqrt{11})}{2 \cdot - 4}$

Move $2$ to the left of $i$ .

$x = \frac{- 14 \pm 2 i \sqrt{11}}{2 \cdot - 4}$

Multiply $2$ by $- 4$ .

$x = \frac{- 14 \pm 2 i \sqrt{11}}{- 8}$

Simplify $\frac{- 14 \pm 2 i \sqrt{11}}{- 8}$ .

$x = \frac{7 \pm i \sqrt{11}}{4}$

Solve Using the Quadratic Formula 13-6x=(2x-5)^2+3

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