Solve 3x^2=8x+9

$$3 x ^ { 2 } = 8 x + 9$$

Solve for x

$$x = \frac{\sqrt{43} + 4}{3} \approx 3.519146175$$

$$x=\frac{4-\sqrt{43}}{3}\approx -0.852479508$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$3x^{2}-8x=9$$

Subtract 9 from both sides.

$$3x^{2}-8x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -8 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\left(-9\right)}}{2\times 3}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 3\left(-9\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-12\left(-9\right)}}{2\times 3}$$

Multiply -12 times -9.

$$x=\frac{-\left(-8\right)±\sqrt{64+108}}{2\times 3}$$

Add 64 to 108.

$$x=\frac{-\left(-8\right)±\sqrt{172}}{2\times 3}$$

Take the square root of 172.

$$x=\frac{-\left(-8\right)±2\sqrt{43}}{2\times 3}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{43}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{8±2\sqrt{43}}{6}$$

Now solve the equation x=\frac{8±2\sqrt{43}}{6} when ± is plus. Add 8 to 2\sqrt{43}.

$$x=\frac{2\sqrt{43}+8}{6}$$

Divide 8+2\sqrt{43} by 6.

$$x=\frac{\sqrt{43}+4}{3}$$

Now solve the equation x=\frac{8±2\sqrt{43}}{6} when ± is minus. Subtract 2\sqrt{43} from 8.

$$x=\frac{8-2\sqrt{43}}{6}$$

Divide 8-2\sqrt{43} by 6.

$$x=\frac{4-\sqrt{43}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{43}+4}{3} x=\frac{4-\sqrt{43}}{3}$$

Steps for Completing the Square

Subtract 8x from both sides.

$$3x^{2}-8x=9$$

Divide both sides by 3.

$$\frac{3x^{2}-8x}{3}=\frac{9}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{8}{3}x=\frac{9}{3}$$

Divide 9 by 3.

$$x^{2}-\frac{8}{3}x=3$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{3}x+\left(-\frac{4}{3}\right)^{2}=3+\left(-\frac{4}{3}\right)^{2}$$

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

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=3+\frac{16}{9}$$

Add 3 to \frac{16}{9}.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{43}{9}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{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}.

$$\left(x-\frac{4}{3}\right)^{2}=\frac{43}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{3}\right)^{2}}=\sqrt{\frac{43}{9}}$$

Simplify.

$$x-\frac{4}{3}=\frac{\sqrt{43}}{3} x-\frac{4}{3}=-\frac{\sqrt{43}}{3}$$

Add \frac{4}{3} to both sides of the equation.

$$x=\frac{\sqrt{43}+4}{3} x=\frac{4-\sqrt{43}}{3}$$