Solve (x) = - 2x ^ 2 - 2x + 9 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$(x)=-2x^{2}-2x+9$$

Solve for x

$x=-3$
$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Add $2x^{2}$ to both sides.

$$x+2x^{2}=-2x+9$$

Add $2x$ to both sides.

$$x+2x^{2}+2x=9$$

Combine $x$ and $2x$ to get $3x$.

$$3x+2x^{2}=9$$

Subtract $9$ from both sides.

$$3x+2x^{2}-9=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2x^{2}+3x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $2x^{2}+ax+bx-9$. To find $a$ and $b$, set up a system to be solved.

$$a+b=3$$ $$ab=2\left(-9\right)=-18$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-18$.

$$-1,18$$ $$-2,9$$ $$-3,6$$

Calculate the sum for each pair.

$$-1+18=17$$ $$-2+9=7$$ $$-3+6=3$$

The solution is the pair that gives sum $3$.

$$a=-3$$ $$b=6$$

Rewrite $2x^{2}+3x-9$ as $\left(2x^{2}-3x\right)+\left(6x-9\right)$.

$$\left(2x^{2}-3x\right)+\left(6x-9\right)$$

Factor out $x$ in the first and $3$ in the second group.

$$x\left(2x-3\right)+3\left(2x-3\right)$$

Factor out common term $2x-3$ by using distributive property.

$$\left(2x-3\right)\left(x+3\right)$$

To find equation solutions, solve $2x-3=0$ and $x+3=0$.

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

Steps Using the Quadratic Formula

Add $2x^{2}$ to both sides.

$$x+2x^{2}=-2x+9$$

Add $2x$ to both sides.

$$x+2x^{2}+2x=9$$

Combine $x$ and $2x$ to get $3x$.

$$3x+2x^{2}=9$$

Subtract $9$ from both sides.

$$3x+2x^{2}-9=0$$

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.

$$2x^{2}+3x-9=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $3$ for $b$, and $-9$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\left(-9\right)}}{2\times 2}$$

Square $3$.

$$x=\frac{-3±\sqrt{9-4\times 2\left(-9\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-3±\sqrt{9-8\left(-9\right)}}{2\times 2}$$

Multiply $-8$ times $-9$.

$$x=\frac{-3±\sqrt{9+72}}{2\times 2}$$

Add $9$ to $72$.

$$x=\frac{-3±\sqrt{81}}{2\times 2}$$

Take the square root of $81$.

$$x=\frac{-3±9}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{-3±9}{4}$$

Now solve the equation $x=\frac{-3±9}{4}$ when $±$ is plus. Add $-3$ to $9$.

$$x=\frac{6}{4}$$

Reduce the fraction $\frac{6}{4}$ to lowest terms by extracting and canceling out $2$.

$$x=\frac{3}{2}$$

Now solve the equation $x=\frac{-3±9}{4}$ when $±$ is minus. Subtract $9$ from $-3$.

$$x=-\frac{12}{4}$$

Divide $-12$ by $4$.

$$x=-3$$

The equation is now solved.

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

Steps for Completing the Square

Add $2x^{2}$ to both sides.

$$x+2x^{2}=-2x+9$$

Add $2x$ to both sides.

$$x+2x^{2}+2x=9$$

Combine $x$ and $2x$ to get $3x$.

$$3x+2x^{2}=9$$

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 $x^{2}+bx=c$.

$$2x^{2}+3x=9$$

Divide both sides by $2$.

$$\frac{2x^{2}+3x}{2}=\frac{9}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$x^{2}+\frac{3}{2}x=\frac{9}{2}$$

Divide $\frac{3}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{3}{4}$. Then add the square of $\frac{3}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=\frac{9}{2}+\left(\frac{3}{4}\right)^{2}$$

Square $\frac{3}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{9}{2}+\frac{9}{16}$$

Add $\frac{9}{2}$ to $\frac{9}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{81}{16}$$

Factor $x^{2}+\frac{3}{2}x+\frac{9}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{9}{4}$$ $$x+\frac{3}{4}=-\frac{9}{4}$$

Subtract $\frac{3}{4}$ from both sides of the equation.

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