completing the square with a coefficient more than 1.

Question

I've tried to solve it, is this right?

$$2x^2+6x+35=0$$

$$2(x^2+6x)+35$$

$$2(x+3)^2+35-9=0$$

$$2(x+3)^2=26=0$$

I was told to write it in the form $a(x+b)^2+c$.

Answer 1

Besides the error saying $2x^2+6x=2(x^2+6x)$, the second line loses the $=0$ at the end, the $-9$ came from inside parentheses multiplied by $2$ so should be $-18$, and the equals sign before the $26$ in the last line should be a plus sign. Finally, the form you are given is an expression while you are working with an equation.

Answer 2

$2x^2+6x+35=2(x^2+3x)+35=2(x+\frac{3}{2})^2+(\frac{61}{2})$

This is what I did to obtain the constant term $\frac{61}{2}$:

Let $d$ be the constant term then $2(\frac{3}{2})^2+d=35 \rightarrow d=35-\frac{9}{2}=\frac{61}{2}$.