How do you solve $- 15 x^{2} - 10 x + 40 = 0$ $-15x^2 - 10x + 40 = 0$ using the quadratic formula?

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How do you solve $- 15 x^{2} - 10 x + 40 = 0$ $-15x^2 - 10x + 40 = 0$ using the quadratic formula?

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Answer 1 · 2015-08-14T20:08:15

$x_{1, 2} = - \frac{1 \pm 5}{3}$ $x_(1,2) = -(1 +-5)/3$

Explanation:

For a general form quadratic equation

$a x^{2} + b x + c = 0$ $color(blue)(ax^2 + bx + c = 0)$

the two possible roots of the equation take the form - this is the quadratic formula

$x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)$

Now, your quadratic equation looks like this

$- 15 x^{2} - 10 x + 40 = 0$ $-15x^2 - 10x + 40 = 0$

You can divide all the terms by $5$ $5$ to simplify it

$- 3 x^{2} - 2 x + 8 = 0$ $-3x^2 - 2x + 8 = 0$

In your case, you have $a = - 3$ $a=-3$ , $b = - 2$ $b=-2$ , and $c = 8$ $c=8$ . THis means that the two solutions will be

$x_{1, 2} = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 \cdot (- 3) \cdot 8}}{2 \cdot (- 3)}$ $x_(1,2) = (-(-2) +- sqrt( (-2)^2 - 4 * (-3) * 8))/(2 * (-3))$

$x_{1, 2} = \frac{2 \pm \sqrt{100}}{- 6}$ $x_(1,2) = (2 +- sqrt(100))/(-6)$

$x_(1,2) = (2 +- 10)/(-6) = -(1 +- 5)/3 = {(x_1 = (-1 - 5)/3 = -2), (x_2 = (-1 + 5)/3 = 4/3) :}$

Your quadratic equation will thus have two roots, $x_1 = color(green)(-2)$ and $x_2 = color(green)(4/3)$ .

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How do you solve $- 15 x^{2} - 10 x + 40 = 0$ $-15x^2 - 10x + 40 = 0$ using the quadratic formula?

1 Answer

Explanation:

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How do you solve -15x^2 - 10x + 40 = 0 −15x2−10x+40=0-15x^2 - 10x + 40 = 0 using the quadratic formula?

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Explanation:

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How do you solve $- 15 x^{2} - 10 x + 40 = 0$ $-15x^2 - 10x + 40 = 0$ using the quadratic formula?