Solve 2x^2+6x-3x-9=1

$$2 x ^ { 2 } + 6 x - 3 x - 9 = 1$$

Solve for x

$$x = \frac{\sqrt{89} - 3}{4} \approx 1.608495283$$

$$x=\frac{-\sqrt{89}-3}{4}\approx -3.108495283$$

Steps Using the Quadratic Formula

Combine 6x and -3x to get 3x.

$$2x^{2}+3x-9=1$$

Subtract 1 from both sides.

$$2x^{2}+3x-9-1=0$$

Subtract 1 from -9 to get -10.

$$2x^{2}+3x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 3 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\left(-10\right)}}{2\times 2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2\left(-10\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8\left(-10\right)}}{2\times 2}$$

Multiply -8 times -10.

$$x=\frac{-3±\sqrt{9+80}}{2\times 2}$$

Add 9 to 80.

$$x=\frac{-3±\sqrt{89}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±\sqrt{89}}{4}$$

Now solve the equation x=\frac{-3±\sqrt{89}}{4} when ± is plus. Add -3 to \sqrt{89}.

$$x=\frac{\sqrt{89}-3}{4}$$

Now solve the equation x=\frac{-3±\sqrt{89}}{4} when ± is minus. Subtract \sqrt{89} from -3.

$$x=\frac{-\sqrt{89}-3}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{89}-3}{4} x=\frac{-\sqrt{89}-3}{4}$$

Steps for Completing the Square

Combine 6x and -3x to get 3x.

$$2x^{2}+3x-9=1$$

Add 9 to both sides.

$$2x^{2}+3x=1+9$$

Add 1 and 9 to get 10.

$$2x^{2}+3x=10$$

Divide both sides by 2.

$$\frac{2x^{2}+3x}{2}=\frac{10}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{3}{2}x=\frac{10}{2}$$

Divide 10 by 2.

$$x^{2}+\frac{3}{2}x=5$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=5+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=5+\frac{9}{16}$$

Add 5 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{89}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{89}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{89}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{89}}{4} x+\frac{3}{4}=-\frac{\sqrt{89}}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{\sqrt{89}-3}{4} x=\frac{-\sqrt{89}-3}{4}$$