(2x)ysquared
(2-r)e-rr
(4 ± i2) where i2 = -1
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}
x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.
(2x)ysquared
(2-r)e-rr
x=4 x=0
(4 ± i2) where i2 = -1
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
x2 + x2 = 2x2
x2+7x+12
x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}
x2 = 6482 = 64x = 8
Equals anything... x is a variable. If that equation was set equal to zero then you could solve for x, but that is not what you have asked.
2x+6x=-9 => 8x=-9=> x=-8/9