Question 110617
Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{7*x^2+2*x-3=0}}} ( notice {{{a=7}}}, {{{b=2}}}, and {{{c=-3}}})





{{{x = (-2 +- sqrt( (2)^2-4*7*-3 ))/(2*7)}}} Plug in a=7, b=2, and c=-3




{{{x = (-2 +- sqrt( 4-4*7*-3 ))/(2*7)}}} Square 2 to get 4  




{{{x = (-2 +- sqrt( 4+84 ))/(2*7)}}} Multiply {{{-4*-3*7}}} to get {{{84}}}




{{{x = (-2 +- sqrt( 88 ))/(2*7)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-2 +- 2*sqrt(22))/(2*7)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (-2 +- 2*sqrt(22))/14}}} Multiply 2 and 7 to get 14


So now the expression breaks down into two parts


{{{x = (-2 + 2*sqrt(22))/14}}} or {{{x = (-2 - 2*sqrt(22))/14}}}



Now break up the fraction



{{{x=-2/14+2*sqrt(22)/14}}} or {{{x=-2/14-2*sqrt(22)/14}}}



Simplify



{{{x=-1 / 7+sqrt(22)/7}}} or {{{x=-1 / 7-sqrt(22)/7}}}



So these expressions approximate to


{{{x=0.527202251403347}}} or {{{x=-0.812916537117633}}}



So our solutions are:

{{{x=0.527202251403347}}} or {{{x=-0.812916537117633}}}


Notice when we graph {{{7*x^2+2*x-3}}}, we get:


{{{ graph( 500, 500, -10.8129165371176, 10.5272022514033, -10.8129165371176, 10.5272022514033,7*x^2+2*x+-3) }}}


when we use the root finder feature on a calculator, we find that {{{x=0.527202251403347}}} and {{{x=-0.812916537117633}}}.So this verifies our answer