Solve 0=3x^2-40x-675

$$0 = 3 x ^ { 2 } - 40 x - 675$$

Solve for x

$$x = \frac{5 \sqrt{97} + 20}{3} \approx 23.08142967$$

$$x=\frac{20-5\sqrt{97}}{3}\approx -9.748096336$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}-40x-675=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -40 for b, and -675 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 3\left(-675\right)}}{2\times 3}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 3\left(-675\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-40\right)±\sqrt{1600-12\left(-675\right)}}{2\times 3}$$

Multiply -12 times -675.

$$x=\frac{-\left(-40\right)±\sqrt{1600+8100}}{2\times 3}$$

Add 1600 to 8100.

$$x=\frac{-\left(-40\right)±\sqrt{9700}}{2\times 3}$$

Take the square root of 9700.

$$x=\frac{-\left(-40\right)±10\sqrt{97}}{2\times 3}$$

The opposite of -40 is 40.

$$x=\frac{40±10\sqrt{97}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{40±10\sqrt{97}}{6}$$

Now solve the equation x=\frac{40±10\sqrt{97}}{6} when ± is plus. Add 40 to 10\sqrt{97}.

$$x=\frac{10\sqrt{97}+40}{6}$$

Divide 40+10\sqrt{97} by 6.

$$x=\frac{5\sqrt{97}+20}{3}$$

Now solve the equation x=\frac{40±10\sqrt{97}}{6} when ± is minus. Subtract 10\sqrt{97} from 40.

$$x=\frac{40-10\sqrt{97}}{6}$$

Divide 40-10\sqrt{97} by 6.

$$x=\frac{20-5\sqrt{97}}{3}$$

The equation is now solved.

$$x=\frac{5\sqrt{97}+20}{3} x=\frac{20-5\sqrt{97}}{3}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}-40x-675=0$$

Add 675 to both sides. Anything plus zero gives itself.

$$3x^{2}-40x=675$$

Divide both sides by 3.

$$\frac{3x^{2}-40x}{3}=\frac{675}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{40}{3}x=\frac{675}{3}$$

Divide 675 by 3.

$$x^{2}-\frac{40}{3}x=225$$

Divide -\frac{40}{3}, the coefficient of the x term, by 2 to get -\frac{20}{3}. Then add the square of -\frac{20}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{40}{3}x+\left(-\frac{20}{3}\right)^{2}=225+\left(-\frac{20}{3}\right)^{2}$$

Square -\frac{20}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{40}{3}x+\frac{400}{9}=225+\frac{400}{9}$$

Add 225 to \frac{400}{9}.

$$x^{2}-\frac{40}{3}x+\frac{400}{9}=\frac{2425}{9}$$

Factor x^{2}-\frac{40}{3}x+\frac{400}{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}.

$$\left(x-\frac{20}{3}\right)^{2}=\frac{2425}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{20}{3}\right)^{2}}=\sqrt{\frac{2425}{9}}$$

Simplify.

$$x-\frac{20}{3}=\frac{5\sqrt{97}}{3} x-\frac{20}{3}=-\frac{5\sqrt{97}}{3}$$

Add \frac{20}{3} to both sides of the equation.

$$x=\frac{5\sqrt{97}+20}{3} x=\frac{20-5\sqrt{97}}{3}$$