Question 476930
Please post questions one at a time. I'll do the first one to get you started.


# 1





{{{15x^2+7x=2}}} Start with the given equation.



{{{15x^2+7x-2=0}}} Get every term to the left side.



Notice that the quadratic {{{15x^2+7x-2}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=15}}}, {{{B=7}}}, and {{{C=-2}}}



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



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



{{{x = (-(7) +- sqrt( (7)^2-4(15)(-2) ))/(2(15))}}} Plug in  {{{A=15}}}, {{{B=7}}}, and {{{C=-2}}}



{{{x = (-7 +- sqrt( 49-4(15)(-2) ))/(2(15))}}} Square {{{7}}} to get {{{49}}}. 



{{{x = (-7 +- sqrt( 49--120 ))/(2(15))}}} Multiply {{{4(15)(-2)}}} to get {{{-120}}}



{{{x = (-7 +- sqrt( 49+120 ))/(2(15))}}} Rewrite {{{sqrt(49--120)}}} as {{{sqrt(49+120)}}}



{{{x = (-7 +- sqrt( 169 ))/(2(15))}}} Add {{{49}}} to {{{120}}} to get {{{169}}}



{{{x = (-7 +- sqrt( 169 ))/(30)}}} Multiply {{{2}}} and {{{15}}} to get {{{30}}}. 



{{{x = (-7 +- 13)/(30)}}} Take the square root of {{{169}}} to get {{{13}}}. 



{{{x = (-7 + 13)/(30)}}} or {{{x = (-7 - 13)/(30)}}} Break up the expression. 



{{{x = (6)/(30)}}} or {{{x =  (-20)/(30)}}} Combine like terms. 



{{{x = 1/5}}} or {{{x = -2/3}}} Simplify. 



So the solutions are {{{x = 1/5}}} or {{{x = -2/3}}}