Question 164512


{{{4x^2+28x-15=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=4}}}, {{{b=28}}}, and {{{c=-15}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(28) +- sqrt( (28)^2-4(4)(-15) ))/(2(4))}}} Plug in  {{{a=4}}}, {{{b=28}}}, and {{{c=-15}}}



{{{x = (-28 +- sqrt( 784-4(4)(-15) ))/(2(4))}}} Square {{{28}}} to get {{{784}}}. 



{{{x = (-28 +- sqrt( 784--240 ))/(2(4))}}} Multiply {{{4(4)(-15)}}} to get {{{-240}}}



{{{x = (-28 +- sqrt( 784+240 ))/(2(4))}}} Rewrite {{{sqrt(784--240)}}} as {{{sqrt(784+240)}}}



{{{x = (-28 +- sqrt( 1024 ))/(2(4))}}} Add {{{784}}} to {{{240}}} to get {{{1024}}}



{{{x = (-28 +- sqrt( 1024 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (-28 +- 32)/(8)}}} Take the square root of {{{1024}}} to get {{{32}}}. 



{{{x = (-28 + 32)/(8)}}} or {{{x = (-28 - 32)/(8)}}} Break up the expression. 



{{{x = (4)/(8)}}} or {{{x =  (-60)/(8)}}} Combine like terms. 



{{{x = 1/2}}} or {{{x = -15/2}}} Simplify. 



So the answers are {{{x = 1/2}}} or {{{x = -15/2}}}