Solve 2x ^ 2 - 14x + 3x + 14 = 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

$$2x^{2}-14x+3x+14=0$$

Solve for x

$x=2$
$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Combine $-14x$ and $3x$ to get $-11x$.

$$2x^{2}-11x+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $2x^{2}+ax+bx+14$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-11$$ $$ab=2\times 14=28$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $28$.

$$-1,-28$$ $$-2,-14$$ $$-4,-7$$

Calculate the sum for each pair.

$$-1-28=-29$$ $$-2-14=-16$$ $$-4-7=-11$$

The solution is the pair that gives sum $-11$.

$$a=-7$$ $$b=-4$$

Rewrite $2x^{2}-11x+14$ as $\left(2x^{2}-7x\right)+\left(-4x+14\right)$.

$$\left(2x^{2}-7x\right)+\left(-4x+14\right)$$

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

$$x\left(2x-7\right)-2\left(2x-7\right)$$

Factor out common term $2x-7$ by using distributive property.

$$\left(2x-7\right)\left(x-2\right)$$

To find equation solutions, solve $2x-7=0$ and $x-2=0$.

$$x=\frac{7}{2}$$ $$x=2$$

Steps Using the Quadratic Formula

Combine $-14x$ and $3x$ to get $-11x$.

$$2x^{2}-11x+14=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-11$ for $b$, and $14$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\times 14}}{2\times 2}$$

Square $-11$.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 2\times 14}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-11\right)±\sqrt{121-8\times 14}}{2\times 2}$$

Multiply $-8$ times $14$.

$$x=\frac{-\left(-11\right)±\sqrt{121-112}}{2\times 2}$$

Add $121$ to $-112$.

$$x=\frac{-\left(-11\right)±\sqrt{9}}{2\times 2}$$

Take the square root of $9$.

$$x=\frac{-\left(-11\right)±3}{2\times 2}$$

The opposite of $-11$ is $11$.

$$x=\frac{11±3}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{11±3}{4}$$

Now solve the equation $x=\frac{11±3}{4}$ when $±$ is plus. Add $11$ to $3$.

$$x=\frac{14}{4}$$

Reduce the fraction $\frac{14}{4}$ to lowest terms by extracting and canceling out $2$.

$$x=\frac{7}{2}$$

Now solve the equation $x=\frac{11±3}{4}$ when $±$ is minus. Subtract $3$ from $11$.

$$x=\frac{8}{4}$$

Divide $8$ by $4$.

$$x=2$$

The equation is now solved.

$$x=\frac{7}{2}$$ $$x=2$$

Steps for Completing the Square

Combine $-14x$ and $3x$ to get $-11x$.

$$2x^{2}-11x+14=0$$

Subtract $14$ from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-11x=-14$$

Divide both sides by $2$.

$$\frac{2x^{2}-11x}{2}=-\frac{14}{2}$$

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

$$x^{2}-\frac{11}{2}x=-\frac{14}{2}$$

Divide $-14$ by $2$.

$$x^{2}-\frac{11}{2}x=-7$$

Divide $-\frac{11}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{11}{4}$. Then add the square of $-\frac{11}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-7+\left(-\frac{11}{4}\right)^{2}$$

Square $-\frac{11}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-7+\frac{121}{16}$$

Add $-7$ to $\frac{121}{16}$.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{9}{16}$$

Factor $x^{2}-\frac{11}{2}x+\frac{121}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{3}{4}$$ $$x-\frac{11}{4}=-\frac{3}{4}$$

Add $\frac{11}{4}$ to both sides of the equation.

$$x=\frac{7}{2}$$ $$x=2$$