No because the discriminant of this quadratic equation is less than zero.
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That just means that it has no real solutions.
The two solutions are:
x = - 1/2 + 1/2 i sqrt(3)
x = - 1/2 - 1/2 i sqrt(3)
x2 + 49 = 0 ∴ x2 = -49 ∴ x = 7i
y = x2 + x = 0 x (X + 1) = 0 x = 0 is one solution x = -1 is the other
You should be able to look at this equation, or use the discriminant and know that there are no real roots.
x2+10x+1 = -12+2x x2+10x-2x+1+12 = 0 x2+8x+13 = 0 Solving by using the quadratic equation formula: x = - 4 - or + the square root of 3
x2 + 7x = 2x2 + 6 x2 - 2x2 + 7x = 6 -x2 +7x - 6 = 0 or alternatively x2 - 7x + 6 = 0
x2 + 49 = 0 ∴ x2 = -49 ∴ x = 7i
x2+x-6=0 * * * * * Perhaps the solution you are looking for is: x2 + x - 6 = (x - 2)(x + 3) = 0; whence, x = 2 or -3.
x2+24x+144 = 9 x2+24x+144-9 = 0 x2+24x+135 = 0 (x+9)(x+15) = 0 x = -9 or x = -15
y = x2 + x = 0 x (X + 1) = 0 x = 0 is one solution x = -1 is the other
x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.
You should be able to look at this equation, or use the discriminant and know that there are no real roots.
x2+10x+1 = -12+2x x2+10x-2x+1+12 = 0 x2+8x+13 = 0 Solving by using the quadratic equation formula: x = - 4 - or + the square root of 3
x2 + 7x = 2x2 + 6 x2 - 2x2 + 7x = 6 -x2 +7x - 6 = 0 or alternatively x2 - 7x + 6 = 0
x2 + 3x - 10 = (x + 5)(x - 2) ⇒ x = -5 or 2
no
-2
x2+10x+1 = -12+2x x2+8x+13 = 0 The solution: x = -4 + the square root of 3 or x = -4 - the square root of 3