How do you solve $x^{2} - 3 x = 40$ $x^2 - 3x = 40$ by quadratic formula?

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Answer 1 · 2016-01-13T16:10:52

The solutions are
$x = 8, x = - 5$ $color(blue)(x=8 , x=-5$

$x^{2} - 3 x - 40 = 0$ $x^2 -3x - 40=0$

The equation is of the form $a x^{2} + b x + c = 0$ $color(blue)(ax^2+bx+c=0$ where:
$a = 1, b = - 3, c = - 40$ $a=1, b=-3, c=-40$

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

$= (-3)^2-(4*1*(-40))$

$= 9 +160=169$

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

$x = (-(-3)+-sqrt(169))/(2*1) = (3 +-13)/2$

$x= (3 +13)/2 = 16/2 = color(blue)(8$

$x= (3 -13)/2 = -10/2 = color(blue)(-5$

How do you solve x^2 - 3x = 40 x2−3x=40x^2 - 3x = 40 by quadratic formula?