Solve 2x^2+9x=78-3x

$$2 { x }^{ 2 } +9x=78-3x$$

Solve for x

$$x=4\sqrt{3}-3\approx 3.92820323$$

$$x=-4\sqrt{3}-3\approx -9.92820323$$

Steps Using the Quadratic Formula

Subtract 78 from both sides.

$$2x^{2}+9x-78=-3x$$

Add 3x to both sides.

$$2x^{2}+9x-78+3x=0$$

Combine 9x and 3x to get 12x.

$$2x^{2}+12x-78=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 12 for b, and -78 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 2\left(-78\right)}}{2\times 2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 2\left(-78\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-12±\sqrt{144-8\left(-78\right)}}{2\times 2}$$

Multiply -8 times -78.

$$x=\frac{-12±\sqrt{144+624}}{2\times 2}$$

Add 144 to 624.

$$x=\frac{-12±\sqrt{768}}{2\times 2}$$

Take the square root of 768.

$$x=\frac{-12±16\sqrt{3}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-12±16\sqrt{3}}{4}$$

Now solve the equation x=\frac{-12±16\sqrt{3}}{4} when ± is plus. Add -12 to 16\sqrt{3}.

$$x=\frac{16\sqrt{3}-12}{4}$$

Divide -12+16\sqrt{3} by 4.

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

Now solve the equation x=\frac{-12±16\sqrt{3}}{4} when ± is minus. Subtract 16\sqrt{3} from -12.

$$x=\frac{-16\sqrt{3}-12}{4}$$

Divide -12-16\sqrt{3} by 4.

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

The equation is now solved.

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

Steps for Completing the Square

Add 3x to both sides.

$$2x^{2}+9x+3x=78$$

Combine 9x and 3x to get 12x.

$$2x^{2}+12x=78$$

Divide both sides by 2.

$$\frac{2x^{2}+12x}{2}=\frac{78}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{12}{2}x=\frac{78}{2}$$

Divide 12 by 2.

$$x^{2}+6x=\frac{78}{2}$$

Divide 78 by 2.

$$x^{2}+6x=39$$

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.

$$x^{2}+6x+3^{2}=39+3^{2}$$

Square 3.

$$x^{2}+6x+9=39+9$$

Add 39 to 9.

$$x^{2}+6x+9=48$$

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}.

$$\left(x+3\right)^{2}=48$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{48}$$

Simplify.

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

Subtract 3 from both sides of the equation.

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