Solve Using the Quadratic Formula x(5x-4)=28

$x (5 x - 4) = 28$

Step 1

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

Simplify the left side.

Simplify $x (5 x - 4)$ .

Simplify by multiplying through.

Apply the distributive property.

$x (5 x) + x \cdot - 4 = 28$

Reorder.

Rewrite using the commutative property of multiplication.

$5 x \cdot x + x \cdot - 4 = 28$

Move $- 4$ to the left of $x$ .

$5 x \cdot x - 4 \cdot x = 28$

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

Move $x$ .

$5 (x \cdot x) - 4 \cdot x = 28$

Multiply $x$ by $x$ .

$5 x^{2} - 4 \cdot x = 28$

$5 x^{2} - 4 x = 28$

Subtract $28$ from both sides of the equation.

$5 x^{2} - 4 x - 28 = 0$

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 $a = 5$ , $b = - 4$ , and $c = - 28$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (5 \cdot - 28)}}{2 \cdot 5}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 5 \cdot - 28}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot - 28$ .

Multiply $- 4$ by $5$ .

$x = \frac{4 \pm \sqrt{16 - 20 \cdot - 28}}{2 \cdot 5}$

Multiply $- 20$ by $- 28$ .

$x = \frac{4 \pm \sqrt{16 + 560}}{2 \cdot 5}$

Add $16$ and $560$ .

$x = \frac{4 \pm \sqrt{576}}{2 \cdot 5}$

Rewrite $576$ as $24^{2}$ .

$x = \frac{4 \pm \sqrt{24^{2}}}{2 \cdot 5}$

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

$x = \frac{4 \pm 24}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{4 \pm 24}{10}$

Simplify $\frac{4 \pm 24}{10}$ .

$x = \frac{2 \pm 12}{5}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{14}{5}, - 2$

Solve Using the Quadratic Formula x(5x-4)=28

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