Solve 2x^2-3x-7=2(x-m)^2+n

$$2 x ^ { 2 } - 3 x - 7 = 2 ( x - m ) ^ { 2 } + n$$

Solve for n

$$n=4mx-3x-2m^{2}-7$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-m\right)^{2}.

$$2x^{2}-3x-7=2\left(x^{2}-2xm+m^{2}\right)+n$$

Use the distributive property to multiply 2 by x^{2}-2xm+m^{2}.

$$2x^{2}-3x-7=2x^{2}-4xm+2m^{2}+n$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}-4xm+2m^{2}+n=2x^{2}-3x-7$$

Subtract 2x^{2} from both sides.

$$-4xm+2m^{2}+n=2x^{2}-3x-7-2x^{2}$$

Combine 2x^{2} and -2x^{2} to get 0.

$$-4xm+2m^{2}+n=-3x-7$$

Add 4xm to both sides.

$$2m^{2}+n=-3x-7+4xm$$

Subtract 2m^{2} from both sides.

$$n=-3x-7+4xm-2m^{2}$$

Solve for m (complex solution)

$$m=\frac{\sqrt{4x^{2}-6x-2n-14}}{2}+x$$

$$m=-\frac{\sqrt{4x^{2}-6x-2n-14}}{2}+x$$

Solve for m

$$m=\frac{\sqrt{4x^{2}-6x-2n-14}}{2}+x$$

$$m=-\frac{\sqrt{4x^{2}-6x-2n-14}}{2}+x\text{, }n\leq 2x^{2}-3x-7$$