How do you solve $3 x^{2} + 7 x - 13 = 0$ $3x^2+7x-13=0$ ?

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How do you solve $3 x^{2} + 7 x - 13 = 0$ $3x^2+7x-13=0$ ?

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Answer 1 · 2017-01-22T17:22:16

$x = \frac{- 7 + \sqrt{205}}{6}$ $color(blue)x=color(blue)((-7 +sqrt(205))/6$ and $x = \frac{- 7 - \sqrt{205}}{6}$ $color(green)(x=(-7 -sqrt(205))/6$

Explanation:

This is a quadratic equation with respect to $x$ $x$ . We will use the quadratic formula:

For an equation of the type
$a x^{2} + b x + c = 0, a \neq 0$ $color(purple)(ax^2 + bx + c = 0, color(red)(a!=0)$

$x = - b \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a}$ $x=-b pmsqrt(b^2 -4ac)/(2a)$
Putting our values in the equation:

$x = - 7 \pm \frac{\sqrt{{(7)}^{2} - 4 (3) (- 13)}}{2 (3)}$ $x= -7 pmsqrt((7)^2-4(3)(-13))/(2(3))$
$x = - 7 \pm \frac{\sqrt{49 + 156}}{6}$ $x= -7pmsqrt(49+156)/6$

$\Rightarrow$ $=>$ $x = \frac{- 7 \pm \sqrt{205}}{6}$ $color(magenta)(x=color(magenta)((-7 pmsqrt(205))/6$

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How do you solve $3 x^{2} + 7 x - 13 = 0$ $3x^2+7x-13=0$ ?

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