Question

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

What is the solution set of the equation using the quadratic formula?

x2 +2x + 10 = 0

Answer 1

`x^2+2x+10=0\ \ \ \ |-10\\\\x^2+2x=-10\\\\x^2+2\cdot x\cdot1=-10\ \ \ \ |+1^2\\\\\underbrace{x^2+2\cdot x\cdot1+1^2}_((a+b)^2=a^2+2ab+b^2)=-10+1^2\\\\(x+1)^2=-9\to x+1=\pm√(-9)\\\\x+1=\pm3i\ \ \ |-1\\\\\boxed{x=-1-3i\ or\ x=-1+3i}\\\\Used\ i=√(-1)`

Quadratic formula:

`ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\x_1=(-b-\sqrt\Delta)/(2a),\ x_2=(-b+\sqrt\Delta)/(2a)\\\\We\ have\ x^2+2x+10=0\\\\a=1;\ b=2;\ c=10\\\\\Delta=2^2-4\cdot1\cdot10=4-40=-36\\\\\sqrt\Delta=√(-36)=√(36)\cdot√(-1)=6i\\\\x_1=(-2-6i)/(2\cdot1)=-1-3i\\\\x_2=(-2+3i)/(2\cdot1)=-1+3i`

Answer 2

x² + 2x + 10 = 0

a = 1, b = 2, c = 10

x =
`\frac{-(b) +/- \sqrt{(b)^(2)  - 4(a)(c) }}{2a}`

=
`\frac{-(2) +/- \sqrt{(2)^(2)  - 4(1)(10) }}{2(1)}`

=
`(-2 +/- √((4 - 40) ))/(2)`

=
`(-2 +/- √((-36)))/(2)`

=
`(-2 +/- (6i))/(2)`

= -1 +/- (3i)

= -1 + 3i, -1 - 3i