Answer:
Step-by-step explanation:
Expand it: [tex]3x^2 + 18x = -10[/tex]
Add 10 to both sides: [tex]3x^2 + 18x + 10 = 0[/tex]
Divide both sides by 3: [tex]x^2 + 6x + \frac{10}{3} = 0[/tex]
Find a, b for [tex](x + a)^2 + b = x^2 + 2x + a^2 + b \overset{!}{=} x^2 + 6x + \frac{10}{3}[/tex]
It’s [tex]a = \frac{6}{2} = 3 \land b = \frac{10}{3} - a^2 = 3 + \frac{1}{3} - 9 = -6 + \frac{1}{3}[/tex]
So, inserting it leads to: [tex](x+3)^2 -6 + \frac{1}{3} = 0[/tex]
Subtract b from both sides: [tex](x+3)^2 = 6-\frac{1}{3}[/tex]
Apply square root to both sides and consider the two solutions: [tex]x + 3 = \pm\sqrt{6-\frac{1}{3}}[/tex]
Subtract 3 from both sides and you get [tex]x = -3 \pm \sqrt{6 - \frac{1}{3}}[/tex]
