Solve Using the Quadratic Formula x(5x+13)=6

$x (5 x + 13) = 6$

Step 1

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

Simplify the left side.

Simplify $x (5 x + 13)$ .

Simplify by multiplying through.

Apply the distributive property.

$x (5 x) + x \cdot 13 = 6$

Reorder.

Rewrite using the commutative property of multiplication.

$5 x \cdot x + x \cdot 13 = 6$

Move $13$ to the left of $x$ .

$5 x \cdot x + 13 \cdot x = 6$

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

Move $x$ .

$5 (x \cdot x) + 13 \cdot x = 6$

Multiply $x$ by $x$ .

$5 x^{2} + 13 \cdot x = 6$

$5 x^{2} + 13 x = 6$

Subtract $6$ from both sides of the equation.

$5 x^{2} + 13 x - 6 = 0$

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 $a = 5$ , $b = 13$ , and $c = - 6$ into the quadratic formula and solve for $x$ .

$\frac{- 13 \pm \sqrt{13^{2} - 4 \cdot (5 \cdot - 6)}}{2 \cdot 5}$

Step 4

Simplify.

Simplify the numerator.

Raise $13$ to the power of $2$ .

$x = \frac{- 13 \pm \sqrt{169 - 4 \cdot 5 \cdot - 6}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot - 6$ .

Multiply $- 4$ by $5$ .

$x = \frac{- 13 \pm \sqrt{169 - 20 \cdot - 6}}{2 \cdot 5}$

Multiply $- 20$ by $- 6$ .

$x = \frac{- 13 \pm \sqrt{169 + 120}}{2 \cdot 5}$

Add $169$ and $120$ .

$x = \frac{- 13 \pm \sqrt{289}}{2 \cdot 5}$

Rewrite $289$ as $17^{2}$ .

$x = \frac{- 13 \pm \sqrt{17^{2}}}{2 \cdot 5}$

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

$x = \frac{- 13 \pm 17}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 13 \pm 17}{10}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{2}{5}, - 3$

Solve Using the Quadratic Formula x(5x+13)=6

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