Question 172163


{{{7a^2+8a+2=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{aa^2+ba+c}}} where {{{a=7}}}, {{{b=8}}}, and {{{c=2}}}



Let's use the quadratic formula to solve for a



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



{{{a = (-(8) +- sqrt( (8)^2-4(7)(2) ))/(2(7))}}} Plug in  {{{a=7}}}, {{{b=8}}}, and {{{c=2}}}



{{{a = (-8 +- sqrt( 64-4(7)(2) ))/(2(7))}}} Square {{{8}}} to get {{{64}}}. 



{{{a = (-8 +- sqrt( 64-56 ))/(2(7))}}} Multiply {{{4(7)(2)}}} to get {{{56}}}



{{{a = (-8 +- sqrt( 8 ))/(2(7))}}} Subtract {{{56}}} from {{{64}}} to get {{{8}}}



{{{a = (-8 +- sqrt( 8 ))/(14)}}} Multiply {{{2}}} and {{{7}}} to get {{{14}}}. 



{{{a = (-8 +- 2*sqrt(2))/(14)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



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



{{{a = (-4+sqrt(2))/(7)}}} or {{{a = (-4-sqrt(2))/(7)}}} Reduce



So the answers are {{{a = (-4+sqrt(2))/(7)}}} or {{{a = (-4-sqrt(2))/(7)}}} 



which approximate to {{{a=-0.369}}} or {{{a=-0.773}}}