How do you find the zeros, real and imaginary, of y=x^2+3x-4 using the quadratic formula?

Mar 28, 2016

The quatratic formula would be $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Explanation:

In this case $a = 1 , b = 3 , c = - 4$
Using the formula we get:
${x}_{1 , 2} = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \cdot 1 \cdot \left(- 4\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{- 3 \pm \sqrt{9 + 16}}{2} = \frac{- 3 \pm 5}{2}$

So: ${x}_{1} = - 4 \mathmr{and} {x}_{2} = 1$

Extra :
This could have been done easier by factoring:
$= \left(x - 1\right) \left(x + 4\right)$