Solve 20(x)^(2)=-3x+13-2x+12 | 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

$$20 { x }^{ 2 } =-3x+13-2x+12$$

Solve for x

$x = -\frac{5}{4} = -1\frac{1}{4} = -1.25$
$x=1$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Combine $-3x$ and $-2x$ to get $-5x$.

$$20x^{2}=-5x+13+12$$

Add $13$ and $12$ to get $25$.

$$20x^{2}=-5x+25$$

Add $5x$ to both sides.

$$20x^{2}+5x=25$$

Subtract $25$ from both sides.

$$20x^{2}+5x-25=0$$

Divide both sides by $5$.

$$4x^{2}+x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $4x^{2}+ax+bx-5$. To find $a$ and $b$, set up a system to be solved.

$$a+b=1$$ $$ab=4\left(-5\right)=-20$$

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 $-20$.

$$-1,20$$ $$-2,10$$ $$-4,5$$

Calculate the sum for each pair.

$$-1+20=19$$ $$-2+10=8$$ $$-4+5=1$$

The solution is the pair that gives sum $1$.

$$a=-4$$ $$b=5$$

Rewrite $4x^{2}+x-5$ as $\left(4x^{2}-4x\right)+\left(5x-5\right)$.

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

Factor out $4x$ in the first and $5$ in the second group.

$$4x\left(x-1\right)+5\left(x-1\right)$$

Factor out common term $x-1$ by using distributive property.

$$\left(x-1\right)\left(4x+5\right)$$

To find equation solutions, solve $x-1=0$ and $4x+5=0$.

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

Steps Using the Quadratic Formula

Combine $-3x$ and $-2x$ to get $-5x$.

$$20x^{2}=-5x+13+12$$

Add $13$ and $12$ to get $25$.

$$20x^{2}=-5x+25$$

Add $5x$ to both sides.

$$20x^{2}+5x=25$$

Subtract $25$ from both sides.

$$20x^{2}+5x-25=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $20$ for $a$, $5$ for $b$, and $-25$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-5±\sqrt{5^{2}-4\times 20\left(-25\right)}}{2\times 20}$$

Square $5$.

$$x=\frac{-5±\sqrt{25-4\times 20\left(-25\right)}}{2\times 20}$$

Multiply $-4$ times $20$.

$$x=\frac{-5±\sqrt{25-80\left(-25\right)}}{2\times 20}$$

Multiply $-80$ times $-25$.

$$x=\frac{-5±\sqrt{25+2000}}{2\times 20}$$

Add $25$ to $2000$.

$$x=\frac{-5±\sqrt{2025}}{2\times 20}$$

Take the square root of $2025$.

$$x=\frac{-5±45}{2\times 20}$$

Multiply $2$ times $20$.

$$x=\frac{-5±45}{40}$$

Now solve the equation $x=\frac{-5±45}{40}$ when $±$ is plus. Add $-5$ to $45$.

$$x=\frac{40}{40}$$

Divide $40$ by $40$.

$$x=1$$

Now solve the equation $x=\frac{-5±45}{40}$ when $±$ is minus. Subtract $45$ from $-5$.

$$x=-\frac{50}{40}$$

Reduce the fraction $\frac{-50}{40}$ to lowest terms by extracting and canceling out $10$.

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

The equation is now solved.

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

Steps for Completing the Square

Combine $-3x$ and $-2x$ to get $-5x$.

$$20x^{2}=-5x+13+12$$

Add $13$ and $12$ to get $25$.

$$20x^{2}=-5x+25$$

Add $5x$ to both sides.

$$20x^{2}+5x=25$$

Divide both sides by $20$.

$$\frac{20x^{2}+5x}{20}=\frac{25}{20}$$

Dividing by $20$ undoes the multiplication by $20$.

$$x^{2}+\frac{5}{20}x=\frac{25}{20}$$

Reduce the fraction $\frac{5}{20}$ to lowest terms by extracting and canceling out $5$.

$$x^{2}+\frac{1}{4}x=\frac{25}{20}$$

Reduce the fraction $\frac{25}{20}$ to lowest terms by extracting and canceling out $5$.

$$x^{2}+\frac{1}{4}x=\frac{5}{4}$$

Divide $\frac{1}{4}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{8}$. Then add the square of $\frac{1}{8}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\frac{5}{4}+\left(\frac{1}{8}\right)^{2}$$

Square $\frac{1}{8}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{5}{4}+\frac{1}{64}$$

Add $\frac{5}{4}$ to $\frac{1}{64}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{81}{64}$$

Factor $x^{2}+\frac{1}{4}x+\frac{1}{64}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{81}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{81}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{9}{8}$$ $$x+\frac{1}{8}=-\frac{9}{8}$$

Subtract $\frac{1}{8}$ from both sides of the equation.

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