How do you solve $x^{2} - 3 x - 3 = - 5$ $x^2 - 3x - 3 = -5$ using the quadratic formula?

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How do you solve $x^{2} - 3 x - 3 = - 5$ $x^2 - 3x - 3 = -5$ using the quadratic formula?

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Answer 1 · 2018-04-24T23:43:12

$x = 2, x = 1$ $x = 2, x = 1$

Explanation:

The quadratic formula is always fun! Let's start with it.
$\frac{- b \pm \sqrt{b^{2} - (4 a c)}}{2 a}$ $(-b+-sqrt(b^2-(4ac)))/(2a)$

The variables, b a, and c represent the coefficients to the terms you possess in the original equation. $a$ $a$ representing the coefficient attached to the $x^{2}$ $x^2$ term, $b$ $b$ representing the coefficient attached to the $x$ $x$ term, and $c$ $c$ representing the constant, which is, in your case, 2.

So, with this in hand, we can rewrite the formula to obtain:
$\frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - (4 (1) (2))}}{2 (1)}$ $(-(-3)+-sqrt((-3)^2-(4(1)(2))))/(2(1))$

So you simplify to obtain
$\frac{3 \pm \sqrt{1}}{2 (1)}$ $(3+-sqrt(1))/(2(1))$

So you get $2$ $2$ and $1$ $1$ from + and - the square root, respectively.

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How do you solve $x^{2} - 3 x - 3 = - 5$ $x^2 - 3x - 3 = -5$ using the quadratic formula?

1 Answer

Explanation:

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How do you solve $x^{2} - 3 x - 3 = - 5$ $x^2 - 3x - 3 = -5$ using the quadratic formula?