-31
x2 - 3x -28 = 0
move the constant
x2 - 3x = 28
take half of the x term (1.5), square it (2.25), and add to both sides
x2 - 3x + 2.25 = 2.25 + 28
The left side is now a perfect square, simplify
(x - 1.5)2 = 30.25
Taking the square root of both sides, you find that
x - 1.5 = ±5.5
So...
X1 = 1.5 + 5.5 = 7
X2 = 1.5 - 5.5 = -4
3x + 7 = -4x + 28 implies 7x + 7 = 28
This is only half of an equation. If -x^2 + 3x +2y = 0, then x^2 -3x = 2y and y = (x^2-3x)/2 Solving for x is the inverse, but difficult without a second equation.
3x+4=-11 3x=-15 x=-5
x + 7y = 33 multiply through by 3: 3x + 21y = 99 Subtract the second equation from the above equation: 23y = 92 So y = 4 Substituting this value of y in the first equation, x + 28 = 33 ie x = 5 The solution is (x,y) = (5,4)
It is a quadratic equation in X.
3x - 1 = 28 3x = 29 x = 29/3 x= (approximately) 9.67
x2 + 3x - 28 = x2 + 7x - 4x - 28 = x(x + 7) - 4(x + 7) = (x + 7)(x - 4)
If you mean: 3x -7 +4x = 28 then the value of x works out as 5
3x + 7 = -4x + 28 implies 7x + 7 = 28
-31
This is only half of an equation. If -x^2 + 3x +2y = 0, then x^2 -3x = 2y and y = (x^2-3x)/2 Solving for x is the inverse, but difficult without a second equation.
If the equation is (2x)^3-3x, then the answer is -985. However, if the equation if 2(x)^3-3x, then the answer is -235.
x^2-3x-28=0...................
Points: (9, 1) and (4, 16) Slope: (1-16)/(9-4) = -3 Equation: y-1 = -3(x-9) => y = -3x+28 Equation: y-16 = -3(x-4) => y = -3x+28
Points: (9, 1) and (4, 16) Slope: (1-16)/(9-4) = -3 Equation: y-1 = -3(x-9) => y = -3x+28 Equation: y-16 = -3(x-4) => y = -3x+28
0.5
3x+4=-11 3x=-15 x=-5