how to solve quadratics?

Question

4x2 - 16 = 0

2x2 = 16

x2 + 5 = 0

x2 + 5x - 6 = 0

x2 - 25 = 0

3x2 = -9x

2x2 - 4x + 1 = 0

3x2 + 2x = 7

x2 + 8x + 9 = -7

Answer 1

I assume you meant:
4x^2 - 16 = 0
Factor into binomial:
(2x + 4)(2x - 4) = 0
Assume either factor = 0
2x + 4 = 0, 2x = - 4, x = -2
or
2x - 4 = 0, 2x = 4, x = 2
Substitute in original
4(+2^2) -16
4(4) -16
16 - 16 = 0
True

Answer 2

4x2 - 16 = 0 change to 4x2 = 16 divide by 4 so x2=4 and x=2 or x=-2

2x2 = 16 divide by 2 and get x2 = 8 so x = 2*SQRT(2) or x = -2*SQRT(2)

x2 + 5 = 0 sorry x2 = -5 would only work for imaginary numbers

x2 + 5x - 6 = 0 factor (x + 6)*(x - 1) so x = 1 and x = -6

x2 - 25 = 0 change to x2 = 25 so x=5 or x=-5

3x2 = -9x divide by 3 to get x2 = -3x divide by x go get x=-3 and also x=0 since both sides are multiples of x

Genral form of quadratic equation is A*X^2 + B*X + C = 0
Solution to this is: X = [-B +/- SQRT(B^2 - 4*A*C)] / [2*A]

2x2 - 4x + 1 = 0 here A=2, B=-4 and C=1
So x = [4 +/- SQRT(16 - 8)]/]2*2] = 1 +/- (1/2)*SQRT(2)

3x2 + 2x = 7 here A=3, B=2 and C=-7
So x = [-2 +/- SQRT(4 + 84)]/[[2*3] = [-1 +/- SQRT(22)]/3

x2 + 8x + 9 = -7 add 7 to both sides so x2 + 8x + 16 = 0
This can be factored to (x + 4)*(x + 4) = 0 so x=-4 ( there is only one solution since the equation is always greater than or equal 0)

These look OK to me, hope there are no mistakes, I am being distracted

Answer 3

bring 16 to the right..
4x^2=16
x^2=16/4
X^24
x=+2 or -2

2x^2=16
x^2=16/2
x^2=8
X=+sqr8 or - sqr8

x^2+5x-6=0
(x+6)(-1)=0
x=-6 or +1


x^2-25=0
bring 25 to the rightside.
x^2=25
x=+5 or -5

3x^2=-9x
divide the left and right hand side by 3x
3x=-3
x=-1

2x^2-4x+1=0
(x-1)(x-1)=0
x=1

3x^2+2x-7=0
use the x=-b+or- sqr (b^2-4ac) / (2a)
x=-1.897 or 1.23

x^2+8x+9+7=0
x^2+8x+16=0
use the x=-b+or- sqr (b^2-4ac) / (2a)