```Question 91106
{{{25w^2 - 25w = -3}}}

{{{25w^2 - 25w +3 =0}}} Add 3 to both sides

Let's use the quadratic formula to solve for w:

{{{aw^2+bw+c=0}}}

the general solution using the quadratic equation is:

{{{w = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}

So lets solve {{{25*w^2-25*w+3=0}}} ( notice {{{a=25}}}, {{{b=-25}}}, and {{{c=3}}})

{{{w = (--25 +- sqrt( (-25)^2-4*25*3 ))/(2*25)}}} Plug in a=25, b=-25, and c=3

{{{w = (25 +- sqrt( (-25)^2-4*25*3 ))/(2*25)}}} Negate -25 to get 25

{{{w = (25 +- sqrt( 625-4*25*3 ))/(2*25)}}} Square -25 to get 625  (note: remember when you square -25, you must square the negative as well. This is because {{{(-25)^2=-25*-25=625}}}.)

{{{w = (25 +- sqrt( 625+-300 ))/(2*25)}}} Multiply {{{-4*3*25}}} to get {{{-300}}}

{{{w = (25 +- sqrt( 325 ))/(2*25)}}} Combine like terms in the radicand (everything under the square root)

{{{w = (25 +- 5*sqrt(13))/(2*25)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)

{{{w = (25 +- 5*sqrt(13))/50}}} Multiply 2 and 25 to get 50

So now the expression breaks down into two parts

{{{w = (25 + 5*sqrt(13))/50}}} or {{{w = (25 - 5*sqrt(13))/50}}}

Now break up the fraction

{{{w=+25/50+5*sqrt(13)/50}}} or {{{w=+25/50-5*sqrt(13)/50}}}

Simplify

{{{w=1 / 2+sqrt(13)/10}}} or {{{w=1 / 2-sqrt(13)/10}}}

So these expressions approximate to

{{{w=0.860555127546399}}} or {{{w=0.139444872453601}}}

So our solutions are:

{{{w=0.860555127546399}}} or {{{w=0.139444872453601}}}

Notice when we graph {{{25*x^2-25*x+3}}} (just replace w with x), we get:

{{{ graph( 500, 500, -9.8605551275464, 10.8605551275464, -9.8605551275464, 10.8605551275464,25*x^2+-25*x+3) }}}

when we use the root finder feature on a calculator, we find that {{{x=0.860555127546399}}} and {{{x=0.139444872453601}}}.So this verifies our answer```