# How do you solve 2x^2=x+28 ?

Aug 3, 2015

Move all values and variables to the left and either factor, solve with the quadratic equation, or solve by completing the square; you could also leave the expression alone and solve by graphing.

#### Explanation:

$2 {x}^{2} = x + 28$
$2 {x}^{2} - x = 28$
$2 {x}^{2} - x - 28 = 0$

Next, try to factor this expression.

$\left(2 x + a\right) \left(x + b\right) = 0$

Where $a \times b = - 28$ and $2 b x + a x = - x$ or $2 b + a = - 1$

The factors of -28 are:
$- \left(a \times b\right)$ or $- \left(b \times a\right)$
$- \left(1 \times 28\right)$
$- \left(2 \times 14\right)$
$- \left(4 \times 7\right)$

Of these factors, only the last pair will allow you to make $2 b + a = - 1$ by setting $b = - 4$ and $a = 7$. But this means we can factor the expression above.

$\left(2 x + a\right) \left(x + b\right) = 0$
$\left(2 x + 7\right) \left(x - 4\right) = 0$ which is only true when

$x = - \frac{7}{2}$ or $x = 4$.

Check your work by plugging these values back into the original expression.

Aug 3, 2015

${x}_{1 , 2} = \frac{1}{4} \pm \frac{15}{4}$

#### Explanation:

You could solve this quadratic equation by completing the square.

To do that, start by getting your quadratic to the general form

$\textcolor{b l u e}{{x}^{2} + \frac{b}{a} x = - \frac{c}{a}}$

by adding $- x$ to both sides of the equation and dividing your terms by $2$

$2 {x}^{2} - x = \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 28$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} - \frac{x}{2} = \frac{28}{2}$

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

Now, you need to add a term to both sides of the equation so that the left side of the equation can be written as the square of a binomial.

The coefficient of the $x$-term will tell what term you need to add. More specifically, divide this coefficient by $2$ and square the result to get

${\left(- \frac{1}{2} \cdot \frac{1}{2}\right)}^{2} = \frac{1}{16}$

Add $\frac{1}{16}$ to both sides of the equation

${x}^{2} - \frac{x}{2} + \frac{1}{16} = 14 + \frac{1}{16}$

The left side of the equation can now be rewritten as

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

You now have

${\left(x - \frac{1}{4}\right)}^{2} = \frac{225}{16}$

Take the square root of both sides of the equation

$\sqrt{{\left(x - \frac{1}{4}\right)}^{2}} = \sqrt{\frac{225}{16}}$

$x - \frac{1}{4} = \pm \frac{15}{4} \implies {x}_{1 , 2} = \frac{1}{4} \pm \frac{15}{4}$

The two solutions to the equation will thus be

${x}_{1} = \frac{1}{4} + \frac{15}{4} = \textcolor{g r e e n}{4}$ and ${x}_{2} = \frac{1}{4} - \frac{15}{4} = \textcolor{g r e e n}{- \frac{7}{2}}$

Aug 4, 2015

Solve $y = 2 {x}^{2} - x - 28 = 0$ (1)
Transformed equation $y ' = {x}^{2} - x - 56 = 0$(2).
The 2 real roots of equation (1) are: $\left(- \frac{7}{2}\right)$ and $\left(\frac{8}{2} = 4\right)$