Advanced algebra question?

x^2 + 3x = -25

This is the important part:

take the coefficient of x in this case 3, and multiply it by (1/2), then square your result. the number that you get will be added to both sides. this will make the right hand side a perfect square.

x^2 + 3x = -25

3*(1/2) = (3/2), (3/2)^2= 9/4 this is how you complete the square

x^2 + 3x + 9/4 = -25 + 9/4

If you have a hard time factoring the left hand side, just take a look at the number you square two steps back 3/2.

( x + 3/2 )^2 = -100/4 + 9/4

( x + 3/2 )^2 = -91/4

I assume that you want to solve the equation? If so, then now we can square root both sides. don't forget that the right hand side gets plus or minus.

√( ( x + 3/2 )^2 ) = ± √( -91/4)

note that on the left hand side the square root cancels the square. and on the right hand side we have a negative under the radical, so the negative comes out as and an "i".

x + 3/2 = ± i √( 91/4)

since there isn't a number that multiplied by itself gives you 91 that will stay under the radical, however, the square root of 4 is 2.

x + 3/2 = ± i √( 91)/ 2

Now let's subtract both sides by 3/2

x = -3/2 ± i √( 91)/ 2

so your zeros=x-intercepts=roots are:

x = -3/2 + i √( 91)/ 2
and
x = -3/2 - i √( 91)/ 2

:)

x^2 + 3x = -25
x^2 + 3x + (3/2)^2 = -25 + 9/4
(x + 3/2)^2 = -91 / 4

No square roots of negative numbers, so the equation has no real solution. If you are interested in complex solutions, then we can go further:

x + 3/2 = +/- sqrt(-91 / 4)
= +/- i * sqrt(91) / 2
x = (-3 +/- i * sqrt(91)) / 2

x^2 + 3x = -25
x^2 + 3x/2 + 3x/2 = -25
x^2 + 3x/2 + 3x/2 + 9/4 = -25 + 9/4
(x^2 + 3x/2) + (3x/2 + 9/4) = -100/4 + 9/4
x(x + 3/2) + 3/2(x + 3/2) = -91/4
(x + 3/2)^2 = -91/4
x + 3/2 = ±√(-91/4)
x + 3/2 = ±√(-1/2^2 * 91)
x + 3/2 = ±2i√91
x = ±(i/2)√91 - 3/2
x = (±i√91 - 3)/2

x^2 +3x +25=0 , x=-3+(9-100)/2=[-3+(91)^1/2 i]/2 &
x=[-3-(91)^1/2 i]/2 where i^2=-1