# How do you solve x^2-3x = 28?

Aug 25, 2016

$x = + 7 \text{ and } x = - 4$

#### Explanation:

Some times you can spot the factors other times not so easy. If you are ever really stuck and the method of solution is not fixed use the formula or complete the square.

$y = a {x}^{2} + b x + c \to x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
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Given:$\text{ } {x}^{2} - 3 x = 28$

Subtract 28 from both sides

${x}^{2} - 3 x - 28 = 0$

Notice that $4 \times \left(- 7\right) = - 28 \text{ and } - 7 + 4 = - 3$

$\implies \left(x - 7\right) \left(x + 4\right) = 0$

So $x = + 7 \text{ and } x = - 4$

$\textcolor{g r e e n}{\text{Look at the graph and you will see these values are}}$$\textcolor{g r e e n}{\text{where the graph crosses the x axis. (where y=0)}}$

Aug 25, 2016

$x = 7 \mathmr{and} - 4$

#### Explanation:

${x}^{2} - 3 x = 28$

Write in the standard quadratic form
${x}^{2} - 3 x - 28 = 0$

Factorise
$\left(x - 7\right) \left(x + 4\right)$=0

For this to be true either $\left(x - 7\right) = 0 \mathmr{and} \left(x + 4\right) = 0$

Aug 25, 2016

$x =$ either $+ 7$ or $- 4$

#### Explanation:

${x}^{2} - 3 x = 28$

${x}^{2} - 3 x - 28 = 0$

Factorise:
$\left(x - 7\right) \left(x + 4\right) = 0$

$\therefore$ either $x - 7 = 0 \to x = 7$
or $x + 4 = 0 \to x = - 4$

Hence: $x =$ either $+ 7$ or $- 4$