Find the Roots (Zeros) (2x-3)(5x+1)=2x+2/5

$(2 x - 3) (5 x + 1) = 2 x + \frac{2}{5}$

Step 1

Simplify $(2 x - 3) (5 x + 1)$ .

Rewrite.

$0 + 0 + (2 x - 3) (5 x + 1) = 2 x + \frac{2}{5}$

Simplify by adding zeros.

$(2 x - 3) (5 x + 1) = 2 x + \frac{2}{5}$

Expand $(2 x - 3) (5 x + 1)$ using the FOIL Method.

Apply the distributive property.

$2 x (5 x + 1) - 3 (5 x + 1) = 2 x + \frac{2}{5}$

Apply the distributive property.

$2 x (5 x) + 2 x \cdot 1 - 3 (5 x + 1) = 2 x + \frac{2}{5}$

Apply the distributive property.

$2 x (5 x) + 2 x \cdot 1 - 3 (5 x) - 3 \cdot 1 = 2 x + \frac{2}{5}$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 \cdot 5 x \cdot x + 2 x \cdot 1 - 3 (5 x) - 3 \cdot 1 = 2 x + \frac{2}{5}$

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

Move $x$ .

$2 \cdot 5 (x \cdot x) + 2 x \cdot 1 - 3 (5 x) - 3 \cdot 1 = 2 x + \frac{2}{5}$

Multiply $x$ by $x$ .

$2 \cdot 5 x^{2} + 2 x \cdot 1 - 3 (5 x) - 3 \cdot 1 = 2 x + \frac{2}{5}$

Multiply $2$ by $5$ .

$10 x^{2} + 2 x \cdot 1 - 3 (5 x) - 3 \cdot 1 = 2 x + \frac{2}{5}$

Multiply $2$ by $1$ .

$10 x^{2} + 2 x - 3 (5 x) - 3 \cdot 1 = 2 x + \frac{2}{5}$

Multiply $5$ by $- 3$ .

$10 x^{2} + 2 x - 15 x - 3 \cdot 1 = 2 x + \frac{2}{5}$

Multiply $- 3$ by $1$ .

$10 x^{2} + 2 x - 15 x - 3 = 2 x + \frac{2}{5}$

Subtract $15 x$ from $2 x$ .

$10 x^{2} - 13 x - 3 = 2 x + \frac{2}{5}$

Step 2

Move all terms containing $x$ to the left side of the equation.

Subtract $2 x$ from both sides of the equation.

$10 x^{2} - 13 x - 3 - 2 x = \frac{2}{5}$

Subtract $2 x$ from $- 13 x$ .

$10 x^{2} - 15 x - 3 = \frac{2}{5}$

Step 3

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

Subtract $\frac{2}{5}$ from both sides of the equation.

$10 x^{2} - 15 x - 3 - \frac{2}{5} = 0$

Simplify $10 x^{2} - 15 x - 3 - \frac{2}{5}$ .

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$10 x^{2} - 15 x - 3 \cdot \frac{5}{5} - \frac{2}{5} = 0$

Combine $- 3$ and $\frac{5}{5}$ .

$10 x^{2} - 15 x + \frac{- 3 \cdot 5}{5} - \frac{2}{5} = 0$

Combine the numerators over the common denominator.

$10 x^{2} - 15 x + \frac{- 3 \cdot 5 - 2}{5} = 0$

Simplify the numerator.

Multiply $- 3$ by $5$ .

$10 x^{2} - 15 x + \frac{- 15 - 2}{5} = 0$

Subtract $2$ from $- 15$ .

$10 x^{2} - 15 x + \frac{- 17}{5} = 0$

Move the negative in front of the fraction.

$10 x^{2} - 15 x - \frac{17}{5} = 0$

Step 4

Multiply through by the least common denominator $5$ , then simplify.

Apply the distributive property.

$5 (10 x^{2}) + 5 (- 15 x) + 5 (- \frac{17}{5}) = 0$

Simplify.

Multiply $10$ by $5$ .

$50 x^{2} + 5 (- 15 x) + 5 (- \frac{17}{5}) = 0$

Multiply $- 15$ by $5$ .

$50 x^{2} - 75 x + 5 (- \frac{17}{5}) = 0$

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{17}{5}$ into the numerator.

$50 x^{2} - 75 x + 5 (\frac{- 17}{5}) = 0$

Cancel the common factor.

$50 x^{2} - 75 x + 5 (\frac{- 17}{5}) = 0$

Rewrite the expression.

$50 x^{2} - 75 x - 17 = 0$

Step 5

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 6

Substitute the values $a = 50$ , $b = - 75$ , and $c = - 17$ into the quadratic formula and solve for $x$ .

$\frac{75 \pm \sqrt{{(- 75)}^{2} - 4 \cdot (50 \cdot - 17)}}{2 \cdot 50}$

Step 7

Simplify.

Simplify the numerator.

Raise $- 75$ to the power of $2$ .

$x = \frac{75 \pm \sqrt{5625 - 4 \cdot 50 \cdot - 17}}{2 \cdot 50}$

Multiply $- 4 \cdot 50 \cdot - 17$ .

Multiply $- 4$ by $50$ .

$x = \frac{75 \pm \sqrt{5625 - 200 \cdot - 17}}{2 \cdot 50}$

Multiply $- 200$ by $- 17$ .

$x = \frac{75 \pm \sqrt{5625 + 3400}}{2 \cdot 50}$

Add $5625$ and $3400$ .

$x = \frac{75 \pm \sqrt{9025}}{2 \cdot 50}$

Rewrite $9025$ as $95^{2}$ .

$x = \frac{75 \pm \sqrt{95^{2}}}{2 \cdot 50}$

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

$x = \frac{75 \pm 95}{2 \cdot 50}$

Multiply $2$ by $50$ .

$x = \frac{75 \pm 95}{100}$

Simplify $\frac{75 \pm 95}{100}$ .

$x = \frac{15 \pm 19}{20}$

Step 8

The final answer is the combination of both solutions.

$x = \frac{17}{10}, - \frac{1}{5}$

Step 9

Find the Roots (Zeros) (2x-3)(5x+1)=2x+2/5

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