How to solve 7x(5+9x)=2

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 7x(5+9x)=2 | Equatix

Question

$$7x(5+9x)=2$$

Solve for x

$x=\frac{\sqrt{1729}}{126}-\frac{5}{18}\approx 0.052232109$
$x=-\frac{\sqrt{1729}}{126}-\frac{5}{18}\approx -0.607787665$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $7x$ by $5+9x$.

$$35x+63x^{2}=2$$

Subtract $2$ from both sides.

$$35x+63x^{2}-2=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$63x^{2}+35x-2=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $63$ for $a$, $35$ for $b$, and $-2$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-35±\sqrt{35^{2}-4\times 63\left(-2\right)}}{2\times 63}$$

Square $35$.

$$x=\frac{-35±\sqrt{1225-4\times 63\left(-2\right)}}{2\times 63}$$

Multiply $-4$ times $63$.

$$x=\frac{-35±\sqrt{1225-252\left(-2\right)}}{2\times 63}$$

Multiply $-252$ times $-2$.

$$x=\frac{-35±\sqrt{1225+504}}{2\times 63}$$

Add $1225$ to $504$.

$$x=\frac{-35±\sqrt{1729}}{2\times 63}$$

Multiply $2$ times $63$.

$$x=\frac{-35±\sqrt{1729}}{126}$$

Now solve the equation $x=\frac{-35±\sqrt{1729}}{126}$ when $±$ is plus. Add $-35$ to $\sqrt{1729}$.

$$x=\frac{\sqrt{1729}-35}{126}$$

Divide $-35+\sqrt{1729}$ by $126$.

$$x=\frac{\sqrt{1729}}{126}-\frac{5}{18}$$

Now solve the equation $x=\frac{-35±\sqrt{1729}}{126}$ when $±$ is minus. Subtract $\sqrt{1729}$ from $-35$.

$$x=\frac{-\sqrt{1729}-35}{126}$$

Divide $-35-\sqrt{1729}$ by $126$.

$$x=-\frac{\sqrt{1729}}{126}-\frac{5}{18}$$

The equation is now solved.

$$x=\frac{\sqrt{1729}}{126}-\frac{5}{18}$$ $$x=-\frac{\sqrt{1729}}{126}-\frac{5}{18}$$

Steps for Completing the Square

Use the distributive property to multiply $7x$ by $5+9x$.

$$35x+63x^{2}=2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$63x^{2}+35x=2$$

Divide both sides by $63$.

$$\frac{63x^{2}+35x}{63}=\frac{2}{63}$$

Dividing by $63$ undoes the multiplication by $63$.

$$x^{2}+\frac{35}{63}x=\frac{2}{63}$$

Reduce the fraction $\frac{35}{63}$ to lowest terms by extracting and canceling out $7$.

$$x^{2}+\frac{5}{9}x=\frac{2}{63}$$

Divide $\frac{5}{9}$, the coefficient of the $x$ term, by $2$ to get $\frac{5}{18}$. Then add the square of $\frac{5}{18}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{9}x+\left(\frac{5}{18}\right)^{2}=\frac{2}{63}+\left(\frac{5}{18}\right)^{2}$$

Square $\frac{5}{18}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{9}x+\frac{25}{324}=\frac{2}{63}+\frac{25}{324}$$

Add $\frac{2}{63}$ to $\frac{25}{324}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{9}x+\frac{25}{324}=\frac{247}{2268}$$

Factor $x^{2}+\frac{5}{9}x+\frac{25}{324}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{5}{18}\right)^{2}=\frac{247}{2268}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{18}\right)^{2}}=\sqrt{\frac{247}{2268}}$$

Simplify.

$$x+\frac{5}{18}=\frac{\sqrt{1729}}{126}$$ $$x+\frac{5}{18}=-\frac{\sqrt{1729}}{126}$$

Subtract $\frac{5}{18}$ from both sides of the equation.

$$x=\frac{\sqrt{1729}}{126}-\frac{5}{18}$$ $$x=-\frac{\sqrt{1729}}{126}-\frac{5}{18}$$

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Solve for x

Steps Using the Quadratic Formula

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Steps Using the Quadratic Formula

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