How do you solve $7 R^{2} - 14 R + 10 = 0$ $7R^2 -14R + 10 = 0$ using the quadratic formula?

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Answer 1 · 2015-10-25T11:13:03

The solutions are
$x = \frac{14 + \sqrt{- 84}}{14}$ $color(blue)(x= (14+sqrt(-84))/14$

$x = \frac{14 - \sqrt{- 84}}{14}$ $color(blue)(x= (14-sqrt(-84))/14$

$7 R^{2} - 14 R + 10 = 0$ $7 R^2 - 14R +10=0$

The equation is of the form $a R^{2} + b R + c = 0$ $color(blue)(aR^2+bR+c=0$ where:

$a = 7, b = - 14, c = 10$ $a=7, b=-14, c=10$

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

$= (-14)^2-(4*7*10)$

$= 196 -280=-84$

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

$x = (-(-14)+-sqrt(-84))/(2*7) = (14+-sqrt(-84))/14$

The solutions are
$color(blue)(x= (14+sqrt(-84))/14$

$color(blue)(x= (14-sqrt(-84))/14$

How do you solve 7R^2 -14R + 10 = 07R2−14R+10=07R^2 -14R + 10 = 0 using the quadratic formula?