Solve 2x^2 + 19x - 145 = 0?

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Solve 2x^2 + 19x - 145 = 0?

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Answer 1 · 2018-03-15T16:23:42

See a solution process below:

Explanation:

We can use the quadratic equation to solve this problem:

The quadratic formula states:

For $a x^{2} + b x + c = 0$ $color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{b^{2} - (4 a c)}}{2 \cdot a}$ $x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$2$ $color(red)(2)$ for $a$ $color(red)(a)$

$19$ $color(blue)(19)$ for $b$ $color(blue)(b)$

$- 145$ $color(green)(-145)$ for $c$ $color(green)(c)$ gives:

$x = \frac{- 19 \pm \sqrt{19^{2} - (4 \cdot 2 \cdot - 145)}}{2 \cdot 2}$ $x = (-color(blue)(19) +- sqrt(color(blue)(19)^2 - (4 * color(red)(2) * color(green)(-145))))/(2 * color(red)(2))$

$x = \frac{- 19 \pm \sqrt{361 - (8 \cdot - 145)}}{4}$ $x = (-19 +- sqrt(361 - (8 * color(green)(-145))))/4$

$x = \frac{- 19 \pm \sqrt{361 - (- 1160)}}{4}$ $x = (-19 +- sqrt(361 - (-1160)))/4$

$x = \frac{- 19 \pm \sqrt{361 + 1160}}{4}$ $x = (-19 +- sqrt(361 + 1160))/4$

$x = \frac{- 19 \pm \sqrt{1521}}{4}$ $x = (-19 +- sqrt(1521))/4$

$x = \frac{- 19 - 39}{4}$ $x = (-19 - 39)/4$ and $x = \frac{- 19 + 39}{4}$ $x = (-19 + 39)/4$

$x = \frac{- 58}{4}$ $x = (-58)/4$ and $x = \frac{20}{4}$ $x = 20/4$

$x = - 14.5$ $x = -14.5$ and $x = 5$ $x = 5$

The Solution Set Is: $x = {- 14.5, 5}$ $x = {-14.5, 5}$

Answer 2 · 2018-03-15T17:03:26

See details below....

Explanation:

$2 x^{2} + 19 x - 145 = 0$ $2x^2+19x-145=0$

Start by factoring the the left side

$(2 x + 29) (x - 5)$ $(2x + 29)(x-5)$

Then set factors equal to $0$ $0$

$2 x + 29 = 0 or x - 5 = 0$ $2x + 29 = 0 or x-5=0$

$2 x = 0 - 29 or x = 0 + 5$ $2x = 0 - 29 or x= 0 + 5$

$2 x = - 29 or x = 5$ $2x = -29 or x=5$

$x = \frac{- 29}{2} or x = 5$ $x = (-29)/2 or x=5$

Answer 3 · 2018-03-15T17:07:17

By using the quadratic formula, we find that x=5 and x=-14.5

Explanation:

The quadratic formula takes an equation that looks like this:

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

And plugs it into a formula that solves for x:

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

Based on our equation, we know the values of a, b, and c:

$a = 2$ $a=2$
$b = 19$ $b=19$
$c = - 145$ $c=-145$

$\frac{- 19 \pm \sqrt{19^{2} - 4 (2 \times - 145)}}{2 (2)}$ $(-19+-sqrt(19^2-4(2xx-145)))/(2(2))$

$\frac{- 19 \pm \sqrt{361 + 1160}}{4} \Rightarrow \frac{- 19 \pm \sqrt{1521}}{4}$ $(-19+-sqrt(361+1160))/4 rArr (-19+-sqrt(1521))/4$

$\frac{- 19 \pm 39}{4} \Rightarrow x = [5, - 14.5]$ $(-19+-39)/4 rArr x=[5, -14.5]$

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Solve 2x^2 + 19x - 145 = 0?

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