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

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

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Answer 1 · 2015-03-27T01:07:02

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

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

Substituting coefficients from the given quadratic we get:

$r_{0} = \frac{10 \pm \sqrt{100 - 96}}{4}$ $r_0 = (10 +- sqrt(100 -96))/4$
$= \frac{10 \pm 2}{4}$ $= (10 +- 2)/4$

$\to r_{0} = 3$ $rarr r_0 = 3$ or $r_{0} = 2$ $r_0 = 2$

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

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