How do you solve $x^2 - 3x - 4 = 0$ using the quadratic formula?

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Answer 1 · 2015-07-22T08:57:55

Identify the coefficients $a$ , $b$ and $c$ and substitute those values into the quadratic formula to find $x=-1$ or $x=4$ .

$x^2-3x-4 = 0$ is of the form $ax^2+bx+c = 0$ , with $a=1$ , $b=-3$ and $c = -4$ .

The quadratic formula gives us solutions:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$=(3+-sqrt(3^2-(4xx1xx-4)))/(2*1)$

$=(3+-sqrt(9+16))/2$

$=(3+-sqrt(25))/2$

$=(3+-sqrt(5^2))/2$

$=(3+-5)/2$

That is $x=-1$ or $x=4$

How do you solve x^2 - 3x - 4 = 0x^2 - 3x - 4 = 0 using the quadratic formula?