Solve -12x+7=5-2x^2

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

Solve for x

$$x=2\sqrt{2}+3\approx 5.828427125$$

$$x=3-2\sqrt{2}\approx 0.171572875$$

Steps Using the Quadratic Formula

Subtract 5 from both sides.

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

Subtract 5 from 7 to get 2.

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

Add 2x^{2} to both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-12\right)±\sqrt{144-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$x=\frac{-\left(-12\right)±\sqrt{144-16}}{2\times 2}$$

Add 144 to -16.

$$x=\frac{-\left(-12\right)±\sqrt{128}}{2\times 2}$$

Take the square root of 128.

$$x=\frac{-\left(-12\right)±8\sqrt{2}}{2\times 2}$$

The opposite of -12 is 12.

$$x=\frac{12±8\sqrt{2}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{12±8\sqrt{2}}{4}$$

Now solve the equation x=\frac{12±8\sqrt{2}}{4} when ± is plus. Add 12 to 8\sqrt{2}.

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

Divide 12+8\sqrt{2} by 4.

$$x=2\sqrt{2}+3$$

Now solve the equation x=\frac{12±8\sqrt{2}}{4} when ± is minus. Subtract 8\sqrt{2} from 12.

$$x=\frac{12-8\sqrt{2}}{4}$$

Divide 12-8\sqrt{2} by 4.

$$x=3-2\sqrt{2}$$

The equation is now solved.

$$x=2\sqrt{2}+3 x=3-2\sqrt{2}$$

Steps for Completing the Square

Add 2x^{2} to both sides.

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

Subtract 7 from both sides.

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

Subtract 7 from 5 to get -2.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by 2.

$$\frac{2x^{2}-12x}{2}=-\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{12}{2}\right)x=-\frac{2}{2}$$

Divide -12 by 2.

$$x^{2}-6x=-\frac{2}{2}$$

Divide -2 by 2.

$$x^{2}-6x=-1$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-1+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-1+9$$

Add -1 to 9.

$$x^{2}-6x+9=8$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=8$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{8}$$

Simplify.

$$x-3=2\sqrt{2} x-3=-2\sqrt{2}$$

Add 3 to both sides of the equation.

$$x=2\sqrt{2}+3 x=3-2\sqrt{2}$$