How do you solve using the quadratic formula $3 x^{2} - 2 x - 5 = 0$ $3x^2-2x-5=0$ ?

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How do you solve using the quadratic formula $3 x^{2} - 2 x - 5 = 0$ $3x^2-2x-5=0$ ?

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Answer 1 · 2018-08-04T07:04:07

$x = \frac{5}{3}$ $x = 5/3$ and $- 1$ $-1$

Explanation:

$3 x^{2} - 2 x - 5 = 0$ $3x^2 - 2x - 5 = 0$

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

We know that $a = 3$ $a = 3$ , $b = - 2$ $b = -2$ , and $c = - 5$ $c = -5$ based on the equation, so let's plug them into the formula and solve:
$x = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 (3) (- 5)}}{2 (3)}$ $x = (-(-2) +- sqrt((-2)^2 - 4(3)(-5)))/(2(3))$

$x = \frac{2 \pm \sqrt{4 + 60}}{6}$ $x = (2 +- sqrt(4 + 60))/6$

$x = \frac{2 \pm \sqrt{64}}{6}$ $x = (2 +- sqrt(64))/6$

$x = \frac{2 \pm 8}{6}$ $x = (2 +- 8)/6$

$x = \frac{10}{6}$ $x = 10/6$ and $- \frac{6}{6}$ $-6/6$

$x = \frac{5}{3}$ $x = 5/3$ and $- 1$ $-1$

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How do you solve using the quadratic formula $3 x^{2} - 2 x - 5 = 0$ $3x^2-2x-5=0$ ?

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How do you solve using the quadratic formula $3 x^{2} - 2 x - 5 = 0$ $3x^2-2x-5=0$ ?