How do you find the roots for x2−14x−32=0?

Question

Jason Olsen · Accepted Answer

Step 1
In an equation of the following form

a x^{2} + b x + c = 0

the method to find the roots is:
1) calculate

Δ = b^{2} - 4 a c

2) if

Δ = 0

there is only one root

x_{0} = \frac{- b}{2 a}

3) if

Δ > 0

there are two roots

x = \frac{- b - \sqrt{Δ}}{2 a}

and

x_{+} = \frac{- b + \sqrt{Δ}}{2 a}

4) if

Δ < 0

there is no real solution
Example:

x^{2} - 14 x - 32 = 0

\to a = 1; b = - 14; c = - 32

\to Δ = {(- 14)}^{2} - 4 \times 1 \times (- 32) = 196 + 128 = 324

Δ > 0

therefore we have two roots:

x_{-} = \frac{14 - \sqrt{324}}{2} = \frac{14 - 18}{2} = - \frac{4}{2} = - 2

x_{+} = \frac{14 + \sqrt{324}}{2} = \frac{14 + 18}{2} = \frac{32}{2} = 16

Let us check the validity of our results:

{(- 2)}^{2} - 14 \times (- 2) - 32 = 4 + 28 - 32 = 0 \to O K

{(16)}^{2} - 14 \times (16) - 32 = 256 - 224 - 32 = 0 \to O K

kumewekwah0 · Accepted Answer

Step 1
There are several methods we could use. Here's one.
Notice the signs work out, we can factor
$x^{2} - 14 x - 32 = (x + 2) (x - 16)$
So, $x^{2} - 14 x - 32 = 0$ if and only if
$(x + 2) (x - 16) = 0$
Thus, we need
$x + 2 = 0$ or $x - 16 = 0$
The solutions are:
$x = - 2, x = 16$

How do you find the roots for x^{2}-14x-32=0?

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