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)
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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