I need question 2 and the 3 different solutions Solve 3x2 8

I need question 2 and the 3 different solutions

Solve. 3x^2 + 8x + 7 = 10 3x^2 + 8x - 3 = 0 X = -b plus - minus Square root b^2 - 4_ac/2a = -8 plus - minus Square root (8)^2 - 4 (3)(-3)/2(3) Solve. x^3 + 125 = 0 (There are 3 distinct solutions) (sum of cubes) Multiply 4 + 3i by its complex conjugate. (4+3i)(4-3i) = 16-12i + 12i - 9:2 = 16

Solution

2. x^3 +125 =0

Use the formula: a^3 +b^3 = (a + b)(a² – ab + b²)

So, x^3 + 5^3 =0

(x+3)(x^2 -5x +25) =0

x = -3

and x^2 -5x +25 =0 Use quadratic root formula

x = ( -b + /- sqrt(b^2 -4ac)/2a

= ( 5 + /- sqrt( 25 - 25*4)/2 = (5 + / - i*sqrt(75) )/2

= 5/2 +/- 5sqrt(5)/2

So, x = -3 , 5/2 + 5sqrt5/2 ,5/2 - 5sqrt5/2

3. ( 4 +3i)

Complex conjugate of a+ib = a-ib

So ( 4-3i

Now multiply both ( 4+3i)( 4-3i) = 16 +9 = 25