Question 195300
{{{2x^2 + 8x -25 = 5}}}
I'll use the method called "completing the square"
First get all constant terms on the right side, then
divide both sides by {{{2}}}
take 1/2 of the coefficient of the {{{x}}} term, 
square it, then add it to both sides. The left side
will automatically become a perfect square.
{{{2x^2 + 8x = 30}}}
{{{x^2 + 4x = 15}}}
{{{x^2 + 4x + (4/2)^2 = 15 + (4/2)^2}}}
{{{x^2 + 4x + 4 = 15 + 4}}}
{{{(x + 2)^2 = 19}}}
Take the square root of both sides
{{{x + 2 = sqrt(19)}}}
{{{x = sqrt(19) - 2}}}
and
{{{x + 2 = -sqrt(19)}}}
{{{x = -sqrt(19) - 2}}}
I'll check one solution:
{{{2x^2 + 8x -25 = 5}}}
{{{2*(-sqrt(19) -2)^2 + 8*(-sqrt(19) - 2) - 25 = 5}}}
{{{2*(19 + 4*sqrt(19) + 4) - 8*sqrt(19) - 16 - 25 = 5}}}
{{{38 + 8*sqrt(19) + 8 - 8*sqrt(19) - 16 - 25 = 5}}}
{{{46 - 41 = 5}}}
{{{5 = 5}}}
OK