Question 199645
# 1




{{{6x^2+7x=5}}} Start with the given equation.



{{{6x^2+7x-5=0}}} Subtract 5 from both sides.



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=6}}}, {{{B=7}}}, and {{{C=-5}}}



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(6)(-5) ))/(2(6))}}} Plug in  {{{A=6}}}, {{{B=7}}}, and {{{C=-5}}}



{{{x = (-7 +- sqrt( 49-4(6)(-5) ))/(2(6))}}} Square {{{7}}} to get {{{49}}}. 



{{{x = (-7 +- sqrt( 49--120 ))/(2(6))}}} Multiply {{{4(6)(-5)}}} to get {{{-120}}}



{{{x = (-7 +- sqrt( 49+120 ))/(2(6))}}} Rewrite {{{sqrt(49--120)}}} as {{{sqrt(49+120)}}}



{{{x = (-7 +- sqrt( 169 ))/(2(6))}}} Add {{{49}}} to {{{120}}} to get {{{169}}}



{{{x = (-7 +- sqrt( 169 ))/(12)}}} Multiply {{{2}}} and {{{6}}} to get {{{12}}}. 



{{{x = (-7 +- 13)/(12)}}} Take the square root of {{{169}}} to get {{{13}}}. 



{{{x = (-7 + 13)/(12)}}} or {{{x = (-7 - 13)/(12)}}} Break up the expression. 



{{{x = (6)/(12)}}} or {{{x =  (-20)/(12)}}} Combine like terms. 



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



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

  


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# 2


{{{12+5x-2x^2=0}}} Start with the given equation.



{{{-2x^2+5x+12=0}}} Rearrange the terms.



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=-2}}}, {{{B=5}}}, and {{{C=12}}}



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



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



{{{x = (-(5) +- sqrt( (5)^2-4(-2)(12) ))/(2(-2))}}} Plug in  {{{A=-2}}}, {{{B=5}}}, and {{{C=12}}}



{{{x = (-5 +- sqrt( 25-4(-2)(12) ))/(2(-2))}}} Square {{{5}}} to get {{{25}}}. 



{{{x = (-5 +- sqrt( 25--96 ))/(2(-2))}}} Multiply {{{4(-2)(12)}}} to get {{{-96}}}



{{{x = (-5 +- sqrt( 25+96 ))/(2(-2))}}} Rewrite {{{sqrt(25--96)}}} as {{{sqrt(25+96)}}}



{{{x = (-5 +- sqrt( 121 ))/(2(-2))}}} Add {{{25}}} to {{{96}}} to get {{{121}}}



{{{x = (-5 +- sqrt( 121 ))/(-4)}}} Multiply {{{2}}} and {{{-2}}} to get {{{-4}}}. 



{{{x = (-5 +- 11)/(-4)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{x = (-5 + 11)/(-4)}}} or {{{x = (-5 - 11)/(-4)}}} Break up the expression. 



{{{x = (6)/(-4)}}} or {{{x =  (-16)/(-4)}}} Combine like terms. 



{{{x = -3/2}}} or {{{x = 4}}} Simplify. 



So the solutions are {{{x = -3/2}}} or {{{x = 4}}} 



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# 3





{{{5x^2+16x-12=0}}} Start with the given equation.



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=5}}}, {{{B=16}}}, and {{{C=-12}}}



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



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



{{{x = (-(16) +- sqrt( (16)^2-4(5)(-12) ))/(2(5))}}} Plug in  {{{A=5}}}, {{{B=16}}}, and {{{C=-12}}}



{{{x = (-16 +- sqrt( 256-4(5)(-12) ))/(2(5))}}} Square {{{16}}} to get {{{256}}}. 



{{{x = (-16 +- sqrt( 256--240 ))/(2(5))}}} Multiply {{{4(5)(-12)}}} to get {{{-240}}}



{{{x = (-16 +- sqrt( 256+240 ))/(2(5))}}} Rewrite {{{sqrt(256--240)}}} as {{{sqrt(256+240)}}}



{{{x = (-16 +- sqrt( 496 ))/(2(5))}}} Add {{{256}}} to {{{240}}} to get {{{496}}}



{{{x = (-16 +- sqrt( 496 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (-16 +- 4*sqrt(31))/(10)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (-16+4*sqrt(31))/(10)}}} or {{{x = (-16-4*sqrt(31))/(10)}}} Break up the expression.  



{{{x = (-8+2*sqrt(31))/(5)}}} or {{{x = (-8-2*sqrt(31))/(5)}}} Reduce. 



So the solutions are {{{x = (-8+2*sqrt(31))/(5)}}} or {{{x = (-8-2*sqrt(31))/(5)}}}



which approximate to {{{x=0.627}}} or {{{x=-3.827}}}