Question 37567
s^2+6s+25= solve for Solution sets
The s^2 indicates there will be 2 solutions.
Besure that the quadratic equation is equal to zero.
Use quadratic formula to solve.
In this case...
s^2+6s+25=0
{{{s=(-b+-sqrt(b^2-4*a*c))/(2*a)}}}
ax^2+bx+c=0
plug in your variables
a=1 b=6 c=25
So...
{{{s=(-6+-sqrt(6^2-4*1*25))/(2*1)}}}
and...
{{{s=(-6+-sqrt(36-100))/(2)}}}
and...
{{{s=(-6+-sqrt(-64))/(2)}}}
The negative number beneith the radical will result in an imaginary expression.
Thus...
{{{s=(-6+- i* sqrt(64))/(2)}}}
So...
{{{s=(-6+- i* 8)/(2)}}}
or...
{{{s=(-6+- 8i)/(2)}}}
Remember, first factor out the 2 in the numerator and then cancel...
{{{s=(2*-3+- 4i)/(2)}}}
and...
{{{s=(-3+- 4i)}}}
So the solution set is...
{-3+4i,-3-4i}

So for the equation...
2x^2-9x=1   
Use the same process first making equation equal to zero.
So...
2x^2-9x=1 (subtract 1 from both sides of the equation then begin)

2x^2-9x-1=0
Again x^2 indicates 2 solutions
Use quadratic formula by plugging in indicated variables 
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=2 b=-9 c=-1 (remember -b means the OPPOSITE of the value of b)
So...
{{{x = (9 +- sqrt( -9^2-4*2*-1 ))/(2*2) }}}
Then...
{{{x = (9 +- sqrt( 81+8 ))/(4) }}}
And...
{{{x = (9 +- sqrt( 89 ))/(4) }}}
89 is aprime number so there is no square root.
So you final result is the solution set...
{{{(9 + sqrt( 89 ))/(4)}}}, {{{(9 - sqrt( 89 ))/(4)}}}