How do you solve the quadratic $3r^2-5=-10$ using any method?

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Answer 1 · 2016-12-02T04:02:08

$r=+-sqrt15/3i$

$3r^2-5=-10$
$color(white)(a^2a)+5color(white)(aa^2a)+5color(white)(aaa)$ Add 5 to both sides

$3r^2=-5$

$(3r^2)/3=(-5)/3color(white)(aaa)$ Divide both sides by 3

$r^2=-5/3$

$sqrt(r^2)=sqrt(-5/3)color(white)(aaa)$ Square root both sides

$r=+-isqrt(5/3)color(white)(aa)$ The negative inside the square root comes out as $i$ .

$r=+-isqrt(5/3)*sqrt(3/3)color(white)(aaa)$ Rationalize the denominator

$r=+-isqrt(15)/3=+-sqrt15/3i$

How do you solve the quadratic 3r^2-5=-103r^2-5=-10 using any method?