Nonreal answer if we put it into the quadratic equation 0=ax^2bx+c x=(-b[+/-](b^2-4ac)^0.5)/2a x=(7[+/-](7^2-4*3*6)^0.5)/(2*3) x=(7[+/-](49-72)^0.5)/6 x=(7[+/-]((-23)^0.5))/6 x=(7[+/-]((-23)^0.5)/6 As you cannot have the square root of a negative number (unless to take iinto account, but lets not go there) the quadratic does not cross the x-axis, so there a no real values that can be found to solve the equation (which i assume you are trying to do).
3x2 + 17x + c = 0, rearranging gives c = -3x2 - 17x
x2 + 7x = 2x2 + 6 x2 - 2x2 + 7x = 6 -x2 +7x - 6 = 0 or alternatively x2 - 7x + 6 = 0
x=-.25.
63828
The math equation of X times 2 - 7x + 12 times 0 equals 2X - 7x.
x2+7x=-2x2+6 3x2+7x-6=0 (3x-2)(x+3)=0 3x-2=0 or x+3=0 x=2/3 or x=-3
3x2 - 7x = -2 3x2 - 7x + 2 = 0 3x2 - 6x - x + 2 = 0 3x(x - 2) - 1(x - 2) = 0 (x - 2)(3x - 1) = 0 so x - 2 = 0 or 3x - 1 = 0 so x = 2 or x = 1/3
3x2 + 17x + c = 0, rearranging gives c = -3x2 - 17x
x2 + 7x = 2x2 + 6 x2 - 2x2 + 7x = 6 -x2 +7x - 6 = 0 or alternatively x2 - 7x + 6 = 0
x2-5-4x2+3x = 0 -3x2+3x-5 = 0 or as 3x2-3x+5 = 0
x=-.25.
63828
x2+7x+12=0 x2+7x=-12 x+Sqaure Root of 7x= 2(SQ Root Symbol)3
7x + 2 = 6x + 2 if and only x = 0.
7x2-10x+3=0 (7x-3)(x-1)=0
The math equation of X times 2 - 7x + 12 times 0 equals 2X - 7x.
It is: 3x2+6x-11 = 0