Solve x ^ 2 - (- 5 + 2) * x + (- 5 * 2) = 0 | 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

$$x^{2}-(-5+2)x+(-5\times2)=0$$

Solve for x

$x=-5$
$x=2$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Add $-5$ and $2$ to get $-3$.

$$x^{2}-\left(-3x\right)-5\times 2=0$$

The opposite of $-3x$ is $3x$.

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

Multiply $-5$ and $2$ to get $-10$.

$$x^{2}+3x-10=0$$

To solve the equation, factor $x^{2}+3x-10$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.

$$a+b=3$$ $$ab=-10$$

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 $-10$.

$$-1,10$$ $$-2,5$$

Calculate the sum for each pair.

$$-1+10=9$$ $$-2+5=3$$

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

$$a=-2$$ $$b=5$$

Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.

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

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

$$x=2$$ $$x=-5$$

Steps Using Factoring By Grouping

Add $-5$ and $2$ to get $-3$.

$$x^{2}-\left(-3x\right)-5\times 2=0$$

The opposite of $-3x$ is $3x$.

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

Multiply $-5$ and $2$ to get $-10$.

$$x^{2}+3x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-10$. To find $a$ and $b$, set up a system to be solved.

$$a+b=3$$ $$ab=1\left(-10\right)=-10$$

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 $-10$.

$$-1,10$$ $$-2,5$$

Calculate the sum for each pair.

$$-1+10=9$$ $$-2+5=3$$

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

$$a=-2$$ $$b=5$$

Rewrite $x^{2}+3x-10$ as $\left(x^{2}-2x\right)+\left(5x-10\right)$.

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

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

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

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

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

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

$$x=2$$ $$x=-5$$

Steps Using the Quadratic Formula

Add $-5$ and $2$ to get $-3$.

$$x^{2}-\left(-3x\right)-5\times 2=0$$

The opposite of $-3x$ is $3x$.

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

Multiply $-5$ and $2$ to get $-10$.

$$x^{2}+3x-10=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $3$ for $b$, and $-10$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-10\right)}}{2}$$

Square $3$.

$$x=\frac{-3±\sqrt{9-4\left(-10\right)}}{2}$$

Multiply $-4$ times $-10$.

$$x=\frac{-3±\sqrt{9+40}}{2}$$

Add $9$ to $40$.

$$x=\frac{-3±\sqrt{49}}{2}$$

Take the square root of $49$.

$$x=\frac{-3±7}{2}$$

Now solve the equation $x=\frac{-3±7}{2}$ when $±$ is plus. Add $-3$ to $7$.

$$x=\frac{4}{2}$$

Divide $4$ by $2$.

$$x=2$$

Now solve the equation $x=\frac{-3±7}{2}$ when $±$ is minus. Subtract $7$ from $-3$.

$$x=-\frac{10}{2}$$

Divide $-10$ by $2$.

$$x=-5$$

The equation is now solved.

$$x=2$$ $$x=-5$$

Steps for Completing the Square

Add $-5$ and $2$ to get $-3$.

$$x^{2}-\left(-3x\right)-5\times 2=0$$

The opposite of $-3x$ is $3x$.

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

Multiply $-5$ and $2$ to get $-10$.

$$x^{2}+3x-10=0$$

Add $10$ to both sides. Anything plus zero gives itself.

$$x^{2}+3x=10$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=10+\left(\frac{3}{2}\right)^{2}$$

Square $\frac{3}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=10+\frac{9}{4}$$

Add $10$ to $\frac{9}{4}$.

$$x^{2}+3x+\frac{9}{4}=\frac{49}{4}$$

Factor $x^{2}+3x+\frac{9}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{7}{2}$$ $$x+\frac{3}{2}=-\frac{7}{2}$$

Subtract $\frac{3}{2}$ from both sides of the equation.

$$x=2$$ $$x=-5$$