Solve for x 2x(10x+8)=-3(x+1)

$2 x (10 x + 8) = - 3 (x + 1)$

Step 1

Simplify $2 x (10 x + 8)$ .

Start simplifying.

$0 + 0 + 2 x (10 x + 8) = - 3 (x + 1)$

Simplify by multiplying through.

Apply the distributive property.

$2 x (10 x) + 2 x \cdot 8 = - 3 (x + 1)$

Simplify the expression.

Rewrite using the commutative property of multiplication.

$2 \cdot 10 x \cdot x + 2 x \cdot 8 = - 3 (x + 1)$

Multiply $8$ by $2$ .

$2 \cdot 10 x \cdot x + 16 x = - 3 (x + 1)$

Simplify each term.

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

Move $x$ .

$2 \cdot 10 (x \cdot x) + 16 x = - 3 (x + 1)$

Multiply $x$ by $x$ .

$2 \cdot 10 x^{2} + 16 x = - 3 (x + 1)$

Multiply $2$ by $10$ .

$20 x^{2} + 16 x = - 3 (x + 1)$

Step 2

Simplify $- 3 (x + 1)$ .

Apply the distributive property.

$20 x^{2} + 16 x = - 3 x - 3 \cdot 1$

Multiply $- 3$ by $1$ .

$20 x^{2} + 16 x = - 3 x - 3$

Step 3

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

Add $3 x$ to both sides of the equation.

$20 x^{2} + 16 x + 3 x = - 3$

Add $16 x$ and $3 x$ .

$20 x^{2} + 19 x = - 3$

Step 4

Add $3$ to both sides of the equation.

$20 x^{2} + 19 x + 3 = 0$

Step 5

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 20 \cdot 3 = 60$ and whose sum is $b = 19$ .

Factor $19$ out of $19 x$ .

$20 x^{2} + 19 (x) + 3 = 0$

Rewrite $19$ as $4$ plus $15$

$20 x^{2} + (4 + 15) x + 3 = 0$

Apply the distributive property.

$20 x^{2} + 4 x + 15 x + 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(20 x^{2} + 4 x) + 15 x + 3 = 0$

Factor out the greatest common factor (GCF) from each group.

$4 x (5 x + 1) + 3 (5 x + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 x + 1$ .

$(5 x + 1) (4 x + 3) = 0$

Step 6

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$5 x + 1 = 0$

$4 x + 3 = 0$

Step 7

Set $5 x + 1$ equal to $0$ and solve for $x$ .

Set $5 x + 1$ equal to $0$ .

$5 x + 1 = 0$

Solve $5 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$5 x = - 1$

Divide each term in $5 x = - 1$ by $5$ and simplify.

Divide each term in $5 x = - 1$ by $5$ .

$\frac{5 x}{5} = \frac{- 1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 1}{5}$

Divide $x$ by $1$ .

$x = \frac{- 1}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{5}$

Step 8

Set $4 x + 3$ equal to $0$ and solve for $x$ .

Set $4 x + 3$ equal to $0$ .

$4 x + 3 = 0$

Solve $4 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$4 x = - 3$

Divide each term in $4 x = - 3$ by $4$ and simplify.

Divide each term in $4 x = - 3$ by $4$ .

$\frac{4 x}{4} = \frac{- 3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 3}{4}$

Divide $x$ by $1$ .

$x = \frac{- 3}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{4}$

Step 9

The final solution is all the values that make $(5 x + 1) (4 x + 3) = 0$ true.

$x = - \frac{1}{5}, - \frac{3}{4}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1}{5}, - \frac{3}{4}$

Decimal Form:

$x = - 0.2, - 0.75$

Solve for x 2x(10x+8)=-3(x+1)

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