Question 597807
{{{-5x^2+5=-2x }}} Start with the given equation


{{{-5x^2+5+2x = 0 }}} Add 2x to both sides.


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


{{{5x^2-2x-5 = 0 }}} Multiply every term by -1 to make the leading coefficient positive.


Notice that the quadratic {{{5x^2-2x-5}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=5}}}, {{{B=-2}}}, 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 = (-(-2) +- sqrt( (-2)^2-4(5)(-5) ))/(2(5))}}} Plug in  {{{A=5}}}, {{{B=-2}}}, and {{{C=-5}}}



{{{x = (2 +- sqrt( (-2)^2-4(5)(-5) ))/(2(5))}}} Negate {{{-2}}} to get {{{2}}}. 



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



{{{x = (2 +- sqrt( 4--100 ))/(2(5))}}} Multiply {{{4(5)(-5)}}} to get {{{-100}}}



{{{x = (2 +- sqrt( 4+100 ))/(2(5))}}} Rewrite {{{sqrt(4--100)}}} as {{{sqrt(4+100)}}}



{{{x = (2 +- sqrt( 104 ))/(2(5))}}} Add {{{4}}} to {{{100}}} to get {{{104}}}



{{{x = (2 +- sqrt( 104 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (2 +- 2*sqrt(26))/(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 = (2+2*sqrt(26))/(10)}}} or {{{x = (2-2*sqrt(26))/(10)}}} Break up the expression.  



{{{x = (1+sqrt(26))/(5)}}} or {{{x = (1-sqrt(26))/(5)}}} Reduce



So the solutions are {{{x = (1+sqrt(26))/(5)}}} or {{{x = (1-sqrt(26))/(5)}}}