How do you solve $3 x^{2} - 4 x - 5 = 0$ $3x^2-4x-5=0$ using the quadratic formula?

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Answer 1 · 2018-04-05T05:38:12

$x = \frac{2 + \sqrt{19}}{3},$ $x=(2+sqrt19)/3,$ $\frac{2 - \sqrt{19}}{3}$ $(2-sqrt19)/3$

Solve:

$3 x^{2} - 4 x - 5 = 0$ $3x^2-4x-5=0$ is a quadratic equation in standard form:

$a x^{2} + b x + c$ $ax^2+bx+c$ ,

where:

$a = 3$ $a=3$ , $b = - 4$ $b=-4$ , $c = - 5$ $c=-5$

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

Plug in the known values.

$x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 \cdot 3 \cdot - 5}}{2 \cdot 3}$ $x=(-(-4)+-sqrt((-4)^2-4*3*-5))/(2*3)$

Simplify.

$x = \frac{4 \pm \sqrt{76}}{6}$ $x=(4+-sqrt76)/6$

Prime factorize $76$ $76$ .

$x = \frac{4 \pm \sqrt{2^{2} \times 19}}{6}$ $x=(4+-sqrt(2^2xx19))/6$

Apply rule: $\sqrt{a^{2}} = a$ $sqrt(a^2)=a$

$x = \frac{4 \pm 2 \sqrt{19}}{6}$ $x=(4+-2sqrt19)/6$

Simplify.

$x = \frac{2 \pm \sqrt{19}}{3}$ $x=(2+-sqrt19)/3$

Solutions for $x$ $x$ .

$x = \frac{2 + \sqrt{19}}{3},$ $x=(2+sqrt19)/3,$ $\frac{2 - \sqrt{19}}{3}$ $(2-sqrt19)/3$

How do you solve 3x2−4x−5=03x^2-4x-5=0 using the quadratic formula?