Question

3x(x + 6) = -10

Answer 1

Answer:

Step-by-step explanation:

Expand it: $3x^2 + 18x = -10$

Add 10 to both sides: $3x^2 + 18x + 10 = 0$

Divide both sides by 3: $x^2 + 6x + \frac{10}{3} = 0$

Find a, b for $(x + a)^2 + b = x^2 + 2x + a^2 + b \overset{!}{=} x^2 + 6x + \frac{10}{3}$

It’s $a = \frac{6}{2} = 3 \land b = \frac{10}{3} - a^2 = 3 + \frac{1}{3} - 9 = -6 + \frac{1}{3}$

So, inserting it leads to: $(x+3)^2 -6 + \frac{1}{3} = 0$

Subtract b from both sides: $(x+3)^2 = 6-\frac{1}{3}$

Apply square root to both sides and consider the two solutions: $x + 3 = \pm\sqrt{6-\frac{1}{3}}$

Subtract 3 from both sides and you get $x = -3 \pm \sqrt{6 - \frac{1}{3}}$