Question 157084


{{{x^2+8x-33=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=8}}}, and {{{c=-33}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(8) +- sqrt( (8)^2-4(1)(-33) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=8}}}, and {{{c=-33}}}



{{{x = (-8 +- sqrt( 64-4(1)(-33) ))/(2(1))}}} Square {{{8}}} to get {{{64}}}. 



{{{x = (-8 +- sqrt( 64--132 ))/(2(1))}}} Multiply {{{4(1)(-33)}}} to get {{{-132}}}



{{{x = (-8 +- sqrt( 64+132 ))/(2(1))}}} Rewrite {{{sqrt(64--132)}}} as {{{sqrt(64+132)}}}



{{{x = (-8 +- sqrt( 196 ))/(2(1))}}} Add {{{64}}} to {{{132}}} to get {{{196}}}



{{{x = (-8 +- sqrt( 196 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-8 +- 14)/(2)}}} Take the square root of {{{196}}} to get {{{14}}}. 



{{{x = (-8 + 14)/(2)}}} or {{{x = (-8 - 14)/(2)}}} Break up the expression. 



{{{x = (6)/(2)}}} or {{{x =  (-22)/(2)}}} Combine like terms. 



{{{x = 3}}} or {{{x = -11}}} Simplify. 



So the answers are {{{x = 3}}} or {{{x = -11}}}