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

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Answer 1 · 2016-03-28T12:00:35

The quatratic formula would be $(-b+-sqrt(b^2-4ac))/(2a)$

In this case $a=1,b=3,c=-4$
Using the formula we get:
$x_(1,2)=(-3+-sqrt(3^2-4*1*(-4)))/(2*1)$

$x_(1,2)=(-3+-sqrt(9+16))/2=(-3+-5)/2$

So: $x_1=-4and x_2=1$

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

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