Question 379042
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Hi,
using the quadratic formula to solve
(r-3)(r+5)=2
r^2 +2r - 15 = 2
r^2 +2r - 17 = 0
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (-2 +- sqrt( 72 ))/(2) }}}
{{{x = (-2 +- sqrt( 72 ))/(2) }}}
{{{x = (-2 +- sqrt( 36*2 ))/(2) }}}
{{{x = (-2 +- 6*sqrt( 2 ))/(2) }}}
{{{x = (-1 +- 3*sqrt( 2 ))}}}