# How do you solve 5x^2-34x-7=0?

Apr 11, 2018

$x = 7$ and $x = - \frac{1}{5}$

#### Explanation:

1) Multiply the coefficient of x by the constant.

$5 \times 7 = 35$

2) Find a pair of numbers that add to make the second term and multiply to make the third. We can easily spot here that only $35$ and $1$ can make $34$

3) Work out which is negative and which is positive if necessary, as the second term is $- 34 x$, the numbers must be $- 35 x$ and $+ x$

Keep the quadratic constant at it's original, we just need to multiply to find the pair of numbers

$5 {x}^{2} - 34 x - 7 \to 5 {x}^{2} + x - 35 x - 7$

4) Factor out the first $2$ terms:

$5 {x}^{2} + x \to x \left(5 x + 1\right)$

5) Factor out the second $2$ terms.

$- 35 x - 7 \to - 7 \left(5 x + 1\right)$

Notice that both brackets are the same

So the first bracket is $\left(5 x + 1\right)$

6) The second bracket is the other $2$ terms.

We have:

$x \left(5 x + 1\right) - 7 \left(5 x + 1\right)$

$\therefore$ The other bracket must be $\left(x - 7\right)$

7) Solve:

$\left(5 x + 1\right) \left(x - 7\right) = 0$

8) Solve each bracket separately:

$5 x + 1 = 0$

$\to 5 x = - 1$

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

$x - 7 = 0$

$\to x = 7$

Therefore these are our two answers.

$x = 7$ and $x = - \frac{1}{5}$

Remember you can always expand the brackets to check the answer