How do you solve $2r^2-10r+12=0$ ?

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Answer 1 · 2015-03-27T01:07:02

$2r^2 - 10r + 12 = 0$
is of the general form for quadratic equations:
$ax^2 + bx + c = 0$ with ( $r$ in place of $x$ )

The general solution for values of $x$ that satisfy this equation is given by the formula:
$(-b +- sqrt(b^2 - 4ac))/(2a)$

Substituting coefficients from the given quadratic we get:

$r_0 = (10 +- sqrt(100 -96))/4$
$= (10 +- 2)/4$

$rarr r_0 = 3$ or $r_0 = 2$

How do you solve 2r^2-10r+12=02r^2-10r+12=0?