Solve (x+1)(2x+3)=4(x)^(2)-22 | 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

$$(x+1)(2x+3)=4 { x }^{ 2 } -22$$

Solve for x

$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$
$x=5$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Use the distributive property to multiply $x+1$ by $2x+3$ and combine like terms.

$$2x^{2}+5x+3=4x^{2}-22$$

Subtract $4x^{2}$ from both sides.

$$2x^{2}+5x+3-4x^{2}=-22$$

Combine $2x^{2}$ and $-4x^{2}$ to get $-2x^{2}$.

$$-2x^{2}+5x+3=-22$$

Add $22$ to both sides.

$$-2x^{2}+5x+3+22=0$$

Add $3$ and $22$ to get $25$.

$$-2x^{2}+5x+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-2x^{2}+ax+bx+25$. To find $a$ and $b$, set up a system to be solved.

$$a+b=5$$ $$ab=-2\times 25=-50$$

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 $-50$.

$$-1,50$$ $$-2,25$$ $$-5,10$$

Calculate the sum for each pair.

$$-1+50=49$$ $$-2+25=23$$ $$-5+10=5$$

The solution is the pair that gives sum $5$.

$$a=10$$ $$b=-5$$

Rewrite $-2x^{2}+5x+25$ as $\left(-2x^{2}+10x\right)+\left(-5x+25\right)$.

$$\left(-2x^{2}+10x\right)+\left(-5x+25\right)$$

Factor out $2x$ in the first and $5$ in the second group.

$$2x\left(-x+5\right)+5\left(-x+5\right)$$

Factor out common term $-x+5$ by using distributive property.

$$\left(-x+5\right)\left(2x+5\right)$$

To find equation solutions, solve $-x+5=0$ and $2x+5=0$.

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

Steps Using the Quadratic Formula

Use the distributive property to multiply $x+1$ by $2x+3$ and combine like terms.

$$2x^{2}+5x+3=4x^{2}-22$$

Subtract $4x^{2}$ from both sides.

$$2x^{2}+5x+3-4x^{2}=-22$$

Combine $2x^{2}$ and $-4x^{2}$ to get $-2x^{2}$.

$$-2x^{2}+5x+3=-22$$

Add $22$ to both sides.

$$-2x^{2}+5x+3+22=0$$

Add $3$ and $22$ to get $25$.

$$-2x^{2}+5x+25=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $5$ for $b$, and $25$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-2\right)\times 25}}{2\left(-2\right)}$$

Square $5$.

$$x=\frac{-5±\sqrt{25-4\left(-2\right)\times 25}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$x=\frac{-5±\sqrt{25+8\times 25}}{2\left(-2\right)}$$

Multiply $8$ times $25$.

$$x=\frac{-5±\sqrt{25+200}}{2\left(-2\right)}$$

Add $25$ to $200$.

$$x=\frac{-5±\sqrt{225}}{2\left(-2\right)}$$

Take the square root of $225$.

$$x=\frac{-5±15}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$x=\frac{-5±15}{-4}$$

Now solve the equation $x=\frac{-5±15}{-4}$ when $±$ is plus. Add $-5$ to $15$.

$$x=\frac{10}{-4}$$

Reduce the fraction $\frac{10}{-4}$ to lowest terms by extracting and canceling out $2$.

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

Now solve the equation $x=\frac{-5±15}{-4}$ when $±$ is minus. Subtract $15$ from $-5$.

$$x=-\frac{20}{-4}$$

Divide $-20$ by $-4$.

$$x=5$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply $x+1$ by $2x+3$ and combine like terms.

$$2x^{2}+5x+3=4x^{2}-22$$

Subtract $4x^{2}$ from both sides.

$$2x^{2}+5x+3-4x^{2}=-22$$

Combine $2x^{2}$ and $-4x^{2}$ to get $-2x^{2}$.

$$-2x^{2}+5x+3=-22$$

Subtract $3$ from both sides.

$$-2x^{2}+5x=-22-3$$

Subtract $3$ from $-22$ to get $-25$.

$$-2x^{2}+5x=-25$$

Divide both sides by $-2$.

$$\frac{-2x^{2}+5x}{-2}=-\frac{25}{-2}$$

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

$$x^{2}+\frac{5}{-2}x=-\frac{25}{-2}$$

Divide $5$ by $-2$.

$$x^{2}-\frac{5}{2}x=-\frac{25}{-2}$$

Divide $-25$ by $-2$.

$$x^{2}-\frac{5}{2}x=\frac{25}{2}$$

Divide $-\frac{5}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{5}{4}$. Then add the square of $-\frac{5}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=\frac{25}{2}+\left(-\frac{5}{4}\right)^{2}$$

Square $-\frac{5}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{25}{2}+\frac{25}{16}$$

Add $\frac{25}{2}$ to $\frac{25}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{225}{16}$$

Factor $x^{2}-\frac{5}{2}x+\frac{25}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{225}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{15}{4}$$ $$x-\frac{5}{4}=-\frac{15}{4}$$

Add $\frac{5}{4}$ to both sides of the equation.

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