Question

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

Answer 1

x² + 2x + 10 = 0

a = 1, b = 2, c = 10

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

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

  = $\frac{-2 +/- \sqrt{(4 - 40) }}{2}$

  = $\frac{-2 +/- \sqrt{(-36)}}{2}$

  = $\frac{-2 +/- (6i)}{2}$

  = -1 +/- (3i)

 = -1 + 3i, -1 - 3i

Answer 2

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

Quadratic formula:

$ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a},\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\We\ have\ x^2+2x+10=0\\\\a=1;\ b=2;\ c=10\\\\\Delta=2^2-4\cdot1\cdot10=4-40=-36\\\\\sqrt\Delta=\sqrt{-36}=\sqrt{36}\cdot\sqrt{-1}=6i\\\\x_1=\dfrac{-2-6i}{2\cdot1}=-1-3i\\\\x_2=\dfrac{-2+3i}{2\cdot1}=-1+3i$