How do you solve $r^{2} - 10 r - 19 = 0$ $r^ { 2} - 10r - 19= 0$ ?

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How do you solve $r^{2} - 10 r - 19 = 0$ $r^ { 2} - 10r - 19= 0$ ?

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Answer 1 · 2017-04-08T01:45:56

$r = \frac{10 \pm \sqrt{176}}{2}$ $r=(10\pmsqrt176)/2$

Explanation:

$r = \frac{- b \pm {\sqrt{b}}^{2} - 4 a c}{2 a}$ $r=(-b\pmsqrtb^2-4ac)/(2a)$
replace variables with numbers with a=1, b=-10, and c=-19
$r = - (- 10) \pm \frac{\sqrt{{(- 10)}^{2} - 4 (1) (- 19)}}{2 (1)}$ $r=-(-10)\pmsqrt((-10)^2-4(1)(-19))/(2(1))$
$r = \frac{10 \pm \sqrt{100 + 76}}{2}$ $r=(10\pmsqrt(100+76))/(2)$
$r = \frac{10 \pm \sqrt{176}}{2}$ $r=(10\pmsqrt176)/(2)$

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How do you solve $r^{2} - 10 r - 19 = 0$ $r^ { 2} - 10r - 19= 0$ ?

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How do you solve r2−10r−19=0r^ { 2} - 10r - 19= 0?

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How do you solve $r^{2} - 10 r - 19 = 0$ $r^ { 2} - 10r - 19= 0$ ?