Solve 2x^2-7=x^2+3x+9

$$2 { x }^{ 2 } -7 = { x }^{ 2 } +3x+9$$

Solve for x

$$x = \frac{\sqrt{73} + 3}{2} \approx 5.772001873$$

$$x=\frac{3-\sqrt{73}}{2}\approx -2.772001873$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$2x^{2}-7-x^{2}=3x+9$$

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

$$x^{2}-7=3x+9$$

Subtract 3x from both sides.

$$x^{2}-7-3x=9$$

Subtract 9 from both sides.

$$x^{2}-7-3x-9=0$$

Subtract 9 from -7 to get -16.

$$x^{2}-16-3x=0$$

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.

$$x^{2}-3x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-16\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$x=\frac{-\left(-3\right)±\sqrt{9+64}}{2}$$

Add 9 to 64.

$$x=\frac{-\left(-3\right)±\sqrt{73}}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{73}}{2}$$

Now solve the equation x=\frac{3±\sqrt{73}}{2} when ± is plus. Add 3 to \sqrt{73}.

$$x=\frac{\sqrt{73}+3}{2}$$

Now solve the equation x=\frac{3±\sqrt{73}}{2} when ± is minus. Subtract \sqrt{73} from 3.

$$x=\frac{3-\sqrt{73}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{73}+3}{2} x=\frac{3-\sqrt{73}}{2}$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$2x^{2}-7-x^{2}=3x+9$$

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

$$x^{2}-7=3x+9$$

Subtract 3x from both sides.

$$x^{2}-7-3x=9$$

Add 7 to both sides.

$$x^{2}-3x=9+7$$

Add 9 and 7 to get 16.

$$x^{2}-3x=16$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=16+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=16+\frac{9}{4}$$

Add 16 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{73}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{73}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{73}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{73}}{2} x-\frac{3}{2}=-\frac{\sqrt{73}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{73}+3}{2} x=\frac{3-\sqrt{73}}{2}$$