Question 189563


{{{4x^2+150x=3000}}} Start with the given equation.



{{{4x^2+150x-3000=0}}} Subtract 3000 from both sides.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=4}}}, {{{b=150}}}, and {{{c=-3000}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(150) +- sqrt( (150)^2-4(4)(-3000) ))/(2(4))}}} Plug in  {{{a=4}}}, {{{b=150}}}, and {{{c=-3000}}}



{{{x = (-150 +- sqrt( 22500-4(4)(-3000) ))/(2(4))}}} Square {{{150}}} to get {{{22500}}}. 



{{{x = (-150 +- sqrt( 22500--48000 ))/(2(4))}}} Multiply {{{4(4)(-3000)}}} to get {{{-48000}}}



{{{x = (-150 +- sqrt( 22500+48000 ))/(2(4))}}} Rewrite {{{sqrt(22500--48000)}}} as {{{sqrt(22500+48000)}}}



{{{x = (-150 +- sqrt( 70500 ))/(2(4))}}} Add {{{22500}}} to {{{48000}}} to get {{{70500}}}



{{{x = (-150 +- sqrt( 70500 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (-150 +- 10*sqrt(705))/(8)}}} Simplify the square root  



{{{x = (-150+10*sqrt(705))/(8)}}} or {{{x = (-150-10*sqrt(705))/(8)}}} Break up the expression.  



{{{x = (-75+5*sqrt(705))/(4)}}} or {{{x = (-75-5*sqrt(705))/(4)}}} Reduce.



So the answers are {{{x = (-75+5*sqrt(705))/(4)}}} or {{{x = (-75-5*sqrt(705))/(4)}}} 



which approximate to {{{x=14.44}}} or {{{x=-51.94}}}