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

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

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Answer 1 · 2015-07-22T08:57:55

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

Explanation:

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

The quadratic formula gives us solutions:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

$= \frac{3 \pm \sqrt{3^{2} - (4 \times 1 \times - 4)}}{2 \cdot 1}$ $=(3+-sqrt(3^2-(4xx1xx-4)))/(2*1)$

$= \frac{3 \pm \sqrt{9 + 16}}{2}$ $=(3+-sqrt(9+16))/2$

$= \frac{3 \pm \sqrt{25}}{2}$ $=(3+-sqrt(25))/2$

$= \frac{3 \pm \sqrt{5^{2}}}{2}$ $=(3+-sqrt(5^2))/2$

$= \frac{3 \pm 5}{2}$ $=(3+-5)/2$

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

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

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