Solve 4x^2=-32x+12

$$4 x ^ { 2 } = - 32 x + 12$$

Solve for x (complex solution)

$$x=\sqrt{19}-4\approx 0.358898944$$

$$x=-\left(\sqrt{19}+4\right)\approx -8.358898944$$

Steps Using the Quadratic Formula

Add 32x to both sides.

$$4x^{2}+32x=12$$

Subtract 12 from both sides.

$$4x^{2}+32x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 32 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-32±\sqrt{32^{2}-4\times 4\left(-12\right)}}{2\times 4}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\times 4\left(-12\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-32±\sqrt{1024-16\left(-12\right)}}{2\times 4}$$

Multiply -16 times -12.

$$x=\frac{-32±\sqrt{1024+192}}{2\times 4}$$

Add 1024 to 192.

$$x=\frac{-32±\sqrt{1216}}{2\times 4}$$

Take the square root of 1216.

$$x=\frac{-32±8\sqrt{19}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-32±8\sqrt{19}}{8}$$

Now solve the equation x=\frac{-32±8\sqrt{19}}{8} when ± is plus. Add -32 to 8\sqrt{19}.

$$x=\frac{8\sqrt{19}-32}{8}$$

Divide -32+8\sqrt{19} by 8.

$$x=\sqrt{19}-4$$

Now solve the equation x=\frac{-32±8\sqrt{19}}{8} when ± is minus. Subtract 8\sqrt{19} from -32.

$$x=\frac{-8\sqrt{19}-32}{8}$$

Divide -32-8\sqrt{19} by 8.

$$x=-\sqrt{19}-4$$

The equation is now solved.

$$x=\sqrt{19}-4 x=-\sqrt{19}-4$$

Steps for Completing the Square

Add 32x to both sides.

$$4x^{2}+32x=12$$

Divide both sides by 4.

$$\frac{4x^{2}+32x}{4}=\frac{12}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{32}{4}x=\frac{12}{4}$$

Divide 32 by 4.

$$x^{2}+8x=\frac{12}{4}$$

Divide 12 by 4.

$$x^{2}+8x=3$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=3+4^{2}$$

Square 4.

$$x^{2}+8x+16=3+16$$

Add 3 to 16.

$$x^{2}+8x+16=19$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=19$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{19}$$

Simplify.

$$x+4=\sqrt{19} x+4=-\sqrt{19}$$

Subtract 4 from both sides of the equation.

$$x=\sqrt{19}-4 x=-\sqrt{19}-4$$

Solve for x

$$x=\sqrt{19}-4\approx 0.358898944$$

$$x=-\sqrt{19}-4\approx -8.358898944$$

Steps Using the Quadratic Formula

Add 32x to both sides.

$$4x^{2}+32x=12$$

Subtract 12 from both sides.

$$4x^{2}+32x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 32 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-32±\sqrt{32^{2}-4\times 4\left(-12\right)}}{2\times 4}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\times 4\left(-12\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-32±\sqrt{1024-16\left(-12\right)}}{2\times 4}$$

Multiply -16 times -12.

$$x=\frac{-32±\sqrt{1024+192}}{2\times 4}$$

Add 1024 to 192.

$$x=\frac{-32±\sqrt{1216}}{2\times 4}$$

Take the square root of 1216.

$$x=\frac{-32±8\sqrt{19}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-32±8\sqrt{19}}{8}$$

Now solve the equation x=\frac{-32±8\sqrt{19}}{8} when ± is plus. Add -32 to 8\sqrt{19}.

$$x=\frac{8\sqrt{19}-32}{8}$$

Divide -32+8\sqrt{19} by 8.

$$x=\sqrt{19}-4$$

Now solve the equation x=\frac{-32±8\sqrt{19}}{8} when ± is minus. Subtract 8\sqrt{19} from -32.

$$x=\frac{-8\sqrt{19}-32}{8}$$

Divide -32-8\sqrt{19} by 8.

$$x=-\sqrt{19}-4$$

The equation is now solved.

$$x=\sqrt{19}-4 x=-\sqrt{19}-4$$

Steps for Completing the Square

Add 32x to both sides.

$$4x^{2}+32x=12$$

Divide both sides by 4.

$$\frac{4x^{2}+32x}{4}=\frac{12}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{32}{4}x=\frac{12}{4}$$

Divide 32 by 4.

$$x^{2}+8x=\frac{12}{4}$$

Divide 12 by 4.

$$x^{2}+8x=3$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=3+4^{2}$$

Square 4.

$$x^{2}+8x+16=3+16$$

Add 3 to 16.

$$x^{2}+8x+16=19$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=19$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{19}$$

Simplify.

$$x+4=\sqrt{19} x+4=-\sqrt{19}$$

Subtract 4 from both sides of the equation.

$$x=\sqrt{19}-4 x=-\sqrt{19}-4$$