# How do you solve x^2+5x-6=0 using the quadratic formula?

Jul 10, 2018

${x}_{1} = 1$ or ${x}_{2} = - 6$

#### Explanation:

Using the formula

${x}_{1 , 2} = - \frac{p}{2} \pm \sqrt{{p}^{2} / 4 - q}$
$p = 5 , q = - 6$

we get

${x}_{1 , 2} = - \frac{5}{2} \pm \sqrt{\frac{25}{4} + 6}$

$\frac{25}{4} + 6 = \frac{25}{4} + \frac{24}{4} = \frac{49}{4}$

so

${x}_{1 , 2} - \frac{5}{2} \pm \sqrt{\frac{49}{4}}$
so we get

${x}_{1} = 1$

${x}_{2} = - 6$

$x = - 6 , 1$

#### Explanation:

Given quadratic equation: ${x}^{2} + 5 x - 6 = 0$

Comparing the above equation with the standard form of quadratic equation: $a {x}^{2} + b x + c = 0$ we get

$a = 1 , b = 5$ & $c = - 6$

The roots of quadratic equation are given as follows

${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \setminus \frac{- 5 \setminus \pm \setminus \sqrt{{5}^{2} - 4 \left(1\right) \left(- 6\right)}}{2 \left(1\right)}$

$= \setminus \frac{- 5 \setminus \pm 7}{2}$

$= - 6 , 1$

Jul 10, 2018

$x = 1 , - 6$

#### Explanation:

${x}^{2} + 5 x - 6 = 0$ is a quadratic equation in standard form:

$y = a {x}^{2} + b x + c$,

where:

$a = 1$, $b = 5$, $c = - 6$

Substitute $0$ for $y$ and solve for $x$.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 \cdot a}$

Plug in the known values.

$x = \frac{- 5 \pm \sqrt{49}}{2}$

Simplify.

$x = \frac{- 5 \pm 7}{2}$

$x = \frac{- 5 + 7}{2}$, $\frac{- 5 - 7}{2}$

Simplify.

$x = \frac{2}{2}$, $x = - \frac{12}{2}$

Simplify.

$x = 1 , - 6$