Question 557653


{{{v^2+3v-54=0}}} Start with the given equation.



Notice that the quadratic {{{v^2+3v-54}}} is in the form of {{{Av^2+Bv+C}}} where {{{A=1}}}, {{{B=3}}}, and {{{C=-54}}}



Let's use the quadratic formula to solve for "v":



{{{v = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{v = (-(3) +- sqrt( (3)^2-4(1)(-54) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=3}}}, and {{{C=-54}}}



{{{v = (-3 +- sqrt( 9-4(1)(-54) ))/(2(1))}}} Square {{{3}}} to get {{{9}}}. 



{{{v = (-3 +- sqrt( 9--216 ))/(2(1))}}} Multiply {{{4(1)(-54)}}} to get {{{-216}}}



{{{v = (-3 +- sqrt( 9+216 ))/(2(1))}}} Rewrite {{{sqrt(9--216)}}} as {{{sqrt(9+216)}}}



{{{v = (-3 +- sqrt( 225 ))/(2(1))}}} Add {{{9}}} to {{{216}}} to get {{{225}}}



{{{v = (-3 +- sqrt( 225 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{v = (-3 +- 15)/(2)}}} Take the square root of {{{225}}} to get {{{15}}}. 



{{{v = (-3 + 15)/(2)}}} or {{{v = (-3 - 15)/(2)}}} Break up the expression. 



{{{v = (12)/(2)}}} or {{{v =  (-18)/(2)}}} Combine like terms. 



{{{v = 6}}} or {{{v = -9}}} Simplify. 



So the solutions are {{{v = 6}}} or {{{v = -9}}} 


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