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

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

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Answer 1 · 2015-09-11T13:09:22

The solutions are
$x = \frac{- 7 + \sqrt{61}}{2}$ $color(blue)(x=(-7+sqrt(61))/2$

$x = \frac{- 7 - \sqrt{61}}{2}$ $color(blue)(x=(-7-sqrt(61))/2$

Explanation:

The equation $x^{2} + 7 x - 3 = 0$ $x^2+7x-3=0$ is of the form $a x^{2} + b x + c = 0$ $color(blue)(ax^2+bx+c=0$ where:
$a = 1, b = 7, c = - 3$ $a=1, b=7, c=-3$

The Discriminant is given by:
$Delta=b^2-4*a*c$

$= (7)^2-(4*1* (-3))$

$= 49 +12= 61$

The solutions are found using the formula
$x=(-b+-sqrtDelta)/(2*a)$

$x = (-7+-sqrt(61))/(2*1) = (-7+-sqrt(61))/2$

The solutions are
$color(blue)(x=(-7+sqrt(61))/2$

$color(blue)(x=(-7-sqrt(61))/2$

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

1 Answer

Explanation:

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