Question #e94e0

#color(blue)(r^2=-14+9r#

Bring everything to the left hand side

#rarrr^2-9r+14=0#

Now, this is a quadratic equation. Solve it by using the Quadratic formula

#color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)#

Where #x# is the variable #r# and #a,band c# are the coefficients of the terms #(a=1,b=-9 and c=14)#

#rarrr=(-(-9)+-sqrt((-9)^2-4(1)(14)))/(2(1))#

#rarrr=(9+-sqrt(81-56))/(2)#

#rarrr=(9+-sqrt(25))/(2)#

#rarrr=(9+-5)/(2)#

Now, there are two solutions for #r#

#color(purple)(r=(9+5)/2=14/2=7)#

#color(violet)(r=(9-5)/2=4/2=2)#

#color(green)( :.r=7# #color(green)(or# #color(green)(2#

Hope this helps..! :)

Re-arrange it into general form first: #ax^2 +bx +c=0#

It does not matter at all what the variable is!

#r^2 -9r +14 =0#

This means: #a = 1" "b = -9" " c =14#

The quadratic formula is: #x = (-b +- sqrt(b^2 -4ac))/(2a)#

Use the values for #a,b,c# above and substitute.

#r = (-(-9) +- sqrt((-9)^2 -4(1)(14)))/(2(1))" "larr# simplify

#r = (9 +- sqrt((81 -56)))/2#

There are two possible answers for #r#

#r = (9+sqrt25)/2 = 7#

#r = (9-sqrt25)/2 = 2#

We could also have solved the original equation by finding the factors.

#"rearrange and equate to zero"#

#rArrr^2-9r+14=0#

#"compare to the standard form " ax^2+bx+c=0#

#"here " a=1, b=-9, c=14#

#rArrr=(-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(rArrr)=(-(-9)+-sqrt((-9)^2-(4xx1xx14)))/2#

#color(white)(rArrr)=(9+-sqrt(81-56))/2=(9+-sqrt25)/2#

#rArrr=(9+5)/2=7" or " r=(9-5)/2=2#