# I Need Someone To Help Me Solve Quadratic Equations, Like Yesterday, I Am Lost And I Have No Money To Pay For Help. A)x2=2x-13=0 B)4x2-4x+3=0 C)x2+12x-64=0 D)2x2-3x-5=0 Please Can Someone Help Me Solve These? I Was Supposed To Turn Them In At Midnight.

First I will start from third equation
X^2+12X-64=0
X^2+X(16-4)-(16*4)=0
X^2+16X-4X-(16*4)=0
Now we will take common
X(X+16) -4(X+16)=0
(X+16)(X-4)=0
X=-16 and X=4 {you can check this... By solving this (X+16)(X-4) = X^2+12X-64}
Now the forth equation
2X^2-3X-5=0
now we will break the middle term
2X^2-(5-2)X-5=0
2X^2 -5X + 2X - 5 = 0
2X^2+2X-5X-5=0
Now we will take common
2X(X+1) - 5(X+1) = 0
(2X-5)(X+1) = 0
2X-5 = 0 , X+1=0
X=5/2 and X=-1

Now second equation
4X^2-4X+3 = 0
4X^2-(6-2)X+3 = 0
4X^2-6x+2X+3 = 0
4X^2+2X-6X+3=0
2X(2X+1) -3(2X+1)=0
(2X-3)(2X+1)=0

I am not able to understand your first equation
Hope this helps
Gud luck

thanked the writer.
Oddman commented
You factored the second equation with factors that give 4x^2-4x-3 = 0. It's a nice try, but the sign of the 3 is wrong.
Toni Flaugh commented
Thank you for all your help,Toni
Ok for b... These are what you need to make your equations equal to zero...( 4x^2)=0 on both those are your solution set.  Do the same to the first part of c and d then add in the third monominal.
thanked the writer.
Toni Flaugh commented
I still do not understand, Toni
drew commented
Wow
ok where do you want me to start
The easiest way to solve a quadratic is by factoring it, but it usually takes some practice to "see" what the factors are. The first one doesn't look like it lends itself to that method, so the other alternative is to use the "quadratic formula." The formula for the solution to a quadratic equation of the form
ax2+bx+c = 0  is
x = (-b ±√(b2 - 4ac))/(2a)

A) X2 + 2X - 13 = 0    (a=1, b=2, c=-13)
X = (-2 ±√(22 - 4*1*(-13))/(2*1) = (-2±√56)/2
= (-2±√(4*14))/2 = (-2±2√14)/2
X = -1±√14    (that's approximately -4.742 or 2.742)

B) 4x2 - 4x + 3 = 0    (a=4, b=-4, c=3)
x = (-(-4) ±√((-4)2 - 4*4*3))/(2*4) = (4 ±√(16-48))/8
= (4±√-32)/8 = (4±√(16*(-2)))/8 = 4(1±√-2)/8
x = (1±I√2)/2    (because I=√(-1))

C) x2 + 12x - 64 = 0    (this one factors nicely)
(x - 4)(x + 16) = 0
x = 4  or  x = -16    (x takes on values that make the factors be zero)

D) 2x2 - 3x - 5 = 0    (this one also factors nicely)
(2x - 5)(x + 1) = 0
x = 5/2  or  x = -1

thanked the writer.
Toni Flaugh commented
Thank you for all of your help, I stayed up all night and figured them out, but you have done a better job than I, thank you, Toni 