Solve c=(2x^2+3x-4)-3(x^2-3x+4)-4

$$c = ( 2 x ^ { 2 } + 3 x - 4 ) - 3 ( x ^ { 2 } - 3 x + 4 ) - 4$$

Solve for c

$$c=-\left(x-10\right)\left(x-2\right)$$

Solution Steps

Use the distributive property to multiply -3 by x^{2}-3x+4.

$$c=2x^{2}+3x-4-3x^{2}+9x-12-4$$

Combine 2x^{2} and -3x^{2} to get -x^{2}.

$$c=-x^{2}+3x-4+9x-12-4$$

Combine 3x and 9x to get 12x.

$$c=-x^{2}+12x-4-12-4$$

Subtract 12 from -4 to get -16.

$$c=-x^{2}+12x-16-4$$

Subtract 4 from -16 to get -20.

$$c=-x^{2}+12x-20$$

Solve for x (complex solution)

$$x=-\sqrt{16-c}+6$$

$$x=\sqrt{16-c}+6$$

Solve for x

$$x=-\sqrt{16-c}+6$$

$$x=\sqrt{16-c}+6\text{, }c\leq 16$$