Question 307202
The phrase "three times the square of a negative number is added to the product of -7 and the number the result is 20." translates to {{{3x^2-7x=20}}} where 'x' is a negative number.



{{{3x^2-7x=20}}} Start with the given equation.



{{{3x^2-7x-20=0}}} Subtract 20 from both sides.



Notice that the quadratic {{{3x^2-7x-20}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=3}}}, {{{B=-7}}}, and {{{C=-20}}}



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(3)(-20) ))/(2(3))}}} Plug in  {{{A=3}}}, {{{B=-7}}}, and {{{C=-20}}}



{{{x = (7 +- sqrt( (-7)^2-4(3)(-20) ))/(2(3))}}} Negate {{{-7}}} to get {{{7}}}. 



{{{x = (7 +- sqrt( 49-4(3)(-20) ))/(2(3))}}} Square {{{-7}}} to get {{{49}}}. 



{{{x = (7 +- sqrt( 49--240 ))/(2(3))}}} Multiply {{{4(3)(-20)}}} to get {{{-240}}}



{{{x = (7 +- sqrt( 49+240 ))/(2(3))}}} Rewrite {{{sqrt(49--240)}}} as {{{sqrt(49+240)}}}



{{{x = (7 +- sqrt( 289 ))/(2(3))}}} Add {{{49}}} to {{{240}}} to get {{{289}}}



{{{x = (7 +- sqrt( 289 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (7 +- 17)/(6)}}} Take the square root of {{{289}}} to get {{{17}}}. 



{{{x = (7 + 17)/(6)}}} or {{{x = (7 - 17)/(6)}}} Break up the expression. 



{{{x = (24)/(6)}}} or {{{x =  (-10)/(6)}}} Combine like terms. 



{{{x = 4}}} or {{{x = -5/3}}} Simplify. 



So the possible solutions are {{{x = 4}}} or {{{x = -5/3}}} 



Keep in mind that 'x' is a negative number. So the only answer is {{{x = -5/3}}}