You can use the quadratic formula for an equation of the form a*x^2 + b*x + c = 0.
x = (-b +/- sqrt(b^2-4*a*c))/(2*a)
In this case:
a = 1
b = -3
c = -28
x = (3 +/- sqrt(9 - 4*(-28)))/2
x = (3 +/- sqrt(121))/2
x = (3 +/- 11)/2
x = 7 or -4
Another option is to attempt to simplify the equation into the form (x-a)(x-b) = 0.
Sometimes you can do this by observation.
For example:
28 = 4*7
7-4 = 3
Thus:
(x-7)(x+4)=0
In this case, a and b are known as the "roots" of the equation, with a = 7 and b = -4.
x: x2 - 81 = 0
y=±√15
16
4n2-25 = 0 4n2=25 n2=6.25 n=sqrt(6.25) n= +2.5 or -2.5
In order to solve this equation you need to know what -28x-2 equals. If you were to assume it was 0 then x = 0.071 rounded
2sinx+1 equals 0
For y - 2y - 3y equals 0, y equals 0.
many solutions
"x equals 0" is an equality, not an inequality. The question is, therefore, not consistent.
Suppose x3-4x = 0. To solve, factor: x3-4x = x(x2-4) = x(x+2)(x-2) = 0 Now, a product equals 0 if and only one or more of the factors equals 0, so set each factor to 0 and solve. The roots are 0,-2 and +2.
x: x2 - 81 = 0
Solve this problem -x squared -40x- 80 =0
x=2
10x=x 9x=0 x=0
Divide both sides by 8: m = 0
x=-.25.
-y + y equals 0.