How do you solve $5 r^{2} = 80$ $5r^2 = 80$ using the quadratic formula?

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How do you solve $5 r^{2} = 80$ $5r^2 = 80$ using the quadratic formula?

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Answer 1 · 2018-04-24T10:17:12

$\pm 4$ $+- 4$

Explanation:

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$
$5 r^{2} + 0 r + (- 80) = 0$ $5r^2+0r+(-80)=0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b +- sqrt(b^2-4ac))/(2a)$
$r = \frac{0 \pm \sqrt{0^{2} - 4 \cdot 5 \cdot - 80}}{2 \cdot 5}$ $r=(0+- sqrt(0^2-4*5*-80))/(2*5)$

$r = \frac{0 \pm \sqrt{0 + 1600}}{10}$ $r=(0+- sqrt(0+1600))/10$
$r = \pm \frac{40}{10}$ $r=+- 40/10$
$r = \pm 4$ $r=+- 4$

Answer 2 · 2018-04-24T10:37:35

$r = \pm 4$ $r=+-4$

Explanation:

$Alternative method to quadratic formula$ $"Alternative method to quadratic formula"$

$divide both sides by 5$ $"divide both sides by 5"$

$\Rightarrow r^{2} = 16$ $rArrr^2=16$

$take the square root of both sides$ $color(blue)"take the square root of both sides"$

$\Rightarrow r = \pm \sqrt{16} \leftarrow note plus or minus$ $rArrr=+-sqrt16larrcolor(blue)"note plus or minus"$

$\Rightarrow r = \pm 4$ $rArrr=+-4$

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How do you solve $5 r^{2} = 80$ $5r^2 = 80$ using the quadratic formula?

2 Answers

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