# How do you use the quadratic formula to find both solutions to the quadratic equation x^2+3x-4=0?

Jun 29, 2015

By replacing the variable coefficients in the general quadratic formual with the coefficients for the terms in the given equation we can get $x = 1$ or $x = - 4$

#### Explanation:

For an equation in the form:
$\textcolor{w h i t e}{\text{XXXX}}$$a {x}^{2} + b x + c = 0$
the quadratic formula gives the solutions as
$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

${x}^{2} + 3 x - 4 = 0$ is in this form with
$\textcolor{w h i t e}{\text{XXXX}}$$a = 1$
$\textcolor{w h i t e}{\text{XXXX}}$$b = 3$ and
$\textcolor{w h i t e}{\text{XXXX}}$$c = \left(- 4\right)$

Replacing $a , b , \mathmr{and} c$ in the quadratic formula, we get
$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \left(1\right) \left(- 4\right)}}{2 \left(1\right)}$

$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- 3 \pm \sqrt{9 + 16}}{2}$

$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- 3 \pm 5}{2}$

$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{2}{2}$ or $x = \frac{- 8}{2}$

$x = 1$ or $x = - 4$

Jun 29, 2015

${x}_{1} = - 4$
${x}_{2} = 1$

#### Explanation:

Having the equation x^2+3x−4=0, we find the ${\Delta}_{x}$:
${\Delta}_{x} = {3}^{2} - 4 \cdot 1 \cdot \left(- 4\right) = 9 + 16 = 25$

Now we find ${x}_{1}$ and ${x}_{2}$ with the quadratic formula .

${x}_{\text{1,2}} = \frac{- 3 \pm \sqrt{{\Delta}_{x}}}{2 \cdot 1} = \frac{- 3 \pm 5}{2}$.

The two solutions are:

${x}_{1} = \frac{- 3 - 5}{2} = - 4$

${x}_{2} = \frac{- 3 + 5}{2} = 1$.