Solve 4x^2-x+16+3x=7x-12-5x

$$4 x ^ { 2 } - x + 16 + 3 x = 7 x - 12 - 5 x$$

Solve for x (complex solution)

$$x=-\sqrt{7}i\approx -0-2.645751311i$$

$$x=\sqrt{7}i\approx 2.645751311i$$

Steps by Finding Square Root

Combine -x and 3x to get 2x.

$$4x^{2}+2x+16=7x-12-5x$$

Combine 7x and -5x to get 2x.

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

Subtract 2x from both sides.

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

Combine 2x and -2x to get 0.

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

Subtract 16 from both sides.

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

Subtract 16 from -12 to get -28.

$$4x^{2}=-28$$

Divide both sides by 4.

$$x^{2}=\frac{-28}{4}$$

Divide -28 by 4 to get -7.

$$x^{2}=-7$$

The equation is now solved.

$$x=\sqrt{7}i x=-\sqrt{7}i$$

Steps Using the Quadratic Formula

Combine -x and 3x to get 2x.

$$4x^{2}+2x+16=7x-12-5x$$

Combine 7x and -5x to get 2x.

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

Subtract 2x from both sides.

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

Combine 2x and -2x to get 0.

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

Add 12 to both sides.

$$4x^{2}+16+12=0$$

Add 16 and 12 to get 28.

$$4x^{2}+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 0 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 4\times 28}}{2\times 4}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 4\times 28}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{0±\sqrt{-16\times 28}}{2\times 4}$$

Multiply -16 times 28.

$$x=\frac{0±\sqrt{-448}}{2\times 4}$$

Take the square root of -448.

$$x=\frac{0±8\sqrt{7}i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{0±8\sqrt{7}i}{8}$$

Now solve the equation x=\frac{0±8\sqrt{7}i}{8} when ± is plus.

$$x=\sqrt{7}i$$

Now solve the equation x=\frac{0±8\sqrt{7}i}{8} when ± is minus.

$$x=-\sqrt{7}i$$

The equation is now solved.

$$x=\sqrt{7}i x=-\sqrt{7}i$$