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

Mar 27, 2015

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

The general solution for values of $x$ that satisfy this equation is given by the formula:
$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substituting coefficients from the given quadratic we get:

${r}_{0} = \frac{10 \pm \sqrt{100 - 96}}{4}$
$= \frac{10 \pm 2}{4}$

$\rightarrow {r}_{0} = 3$ or ${r}_{0} = 2$