x2 +2x+2=0 cannot be factored.
If we look at b2 -4ac which is the
we see it is 4-4(2)=4-8 which is negative.
This means there two roots and they are not real.
The quadratic equation tells us the solution (-2+ 2i)/2 and (-2-2i)/2
or -1+i and -1-i.
Use the quadratic formula. x = -4.265564 or -0.234436
(x - 3)(x + 5)
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
2x + 52 = 7 2x + 25 = 7 2x = 7 - 25 2x = -18 x = -18/2 x = -9
(2x + 1)(x + 2)
Add the x's squared together to get 2x^2+x. You cannot add anymore so the answer is 2x^2+x
Use the quadratic formula. x = -4.265564 or -0.234436
(x - 3)(x + 5)
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
2x + 52 = 7 2x + 25 = 7 2x = 7 - 25 2x = -18 x = -18/2 x = -9
(2x + 1)(x + 2)
(5x 2 - 8x + 1) + (2x 2 - 4x - 11)
The solution to the equation 2x plus 2 is 2(x + 1).
x^2+x^2=2x^2
x^2+2x^2=3x^2
2x(x^2+x-6)
(x - 1)(2x^2-1)