Question 219573
{{{4x^2+19x+21=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=4}}}, {{{b=19}}}, and {{{c=21}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(19) +- sqrt( (19)^2-4(4)(21) ))/(2(4))}}} Plug in  {{{a=4}}}, {{{b=19}}}, and {{{c=21}}}



{{{x = (-19 +- sqrt( 361-4(4)(21) ))/(2(4))}}} Square {{{19}}} to get {{{361}}}. 



{{{x = (-19 +- sqrt( 361-336 ))/(2(4))}}} Multiply {{{4(4)(21)}}} to get {{{336}}}



{{{x = (-19 +- sqrt( 25 ))/(2(4))}}} Subtract {{{336}}} from {{{361}}} to get {{{25}}}



{{{x = (-19 +- sqrt( 25 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (-19 +- 5)/(8)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{x = (-19 + 5)/(8)}}} or {{{x = (-19 - 5)/(8)}}} Break up the expression. 



{{{x = (-14)/(8)}}} or {{{x =  (-24)/(8)}}} Combine like terms. 



{{{x = -7/4}}} or {{{x = -3}}} Simplify. 



So the answers are {{{x = -7/4}}} or {{{x = -3}}}