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

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Answer 1 · 2015-07-06T15:24:16

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

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

$x^{2} + 7 x - 3 = 0$ $x^2+7x-3=0$

The equation 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:
$x = color(blue)( (-7+sqrt(61))/(2$

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

How do you solve the equation: x^2 + 7x - 3 = 0x2+7x−3=0x^2 + 7x - 3 = 0?