It doesn't factor neatly. If you solve it, you get the square root of 12/7 which is either 2 times the square root of 3/7 or -2 times the square root of 3/7
yes
Try to factor the left side and set each of the linear expressions equal to zero. 7x^2 +8x -12= 0 (7X-6)(X+2)=0 7X-6=0 X=6/7 X+2=0 X=-2
(x - 6)(x - 2)
(7x - 4)(x + 1 ) making x = either 4/7 or -1 Check: (7 x 16/49) + 12/7 = 16/7 + 12/7 = 4 ok 7 + (-3) = 4 ok...and there you have it.
Using the quadratic equation formula:- x = -4.706950048 and x = 0.849071913
if the x in "7x2" is variable x, then 7x(2)-5x=0 14x-5x=0 9x=0 x=0 if the x in "7x2" is a symbol for multiplication, then 7(2)-5x=0 14-5x=0 -5x= -14 5x=14 x= 14/5 x= 2 4/5 or, to factor it out... x(14-5)=0 x(9)=0 9x=0
yes
7X^2 -56X + 112 = 0 divide through by 7 X^2 - 8X + 16 = 0 factor (X-4)(X-4) X = 4
(x+3)(x-4)
7x2-10x+3=0 (7x-3)(x-1)=0
Try to factor the left side and set each of the linear expressions equal to zero. 7x^2 +8x -12= 0 (7X-6)(X+2)=0 7X-6=0 X=6/7 X+2=0 X=-2
(x + 3)(x + 4)
(x - 6)(x - 2)
(7x - 4)(x + 1 ) making x = either 4/7 or -1 Check: (7 x 16/49) + 12/7 = 16/7 + 12/7 = 4 ok 7 + (-3) = 4 ok...and there you have it.
The "degree" is the highest power - in this case, the 3 in 4x3 (4 times to the third power).
Yes.
a(-14-a) = 0 so a = 0 or a = -14