Solve 0=3w^2+8w-42

$$0 = 3 w ^ { 2 } + 8 w - 42$$

Solve for w

$$w = \frac{\sqrt{142} - 4}{3} \approx 2.638791763$$

$$w=\frac{-\sqrt{142}-4}{3}\approx -5.305458429$$

Steps Using the Quadratic Formula

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

$$3w^{2}+8w-42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 8 for b, and -42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$w=\frac{-8±\sqrt{8^{2}-4\times 3\left(-42\right)}}{2\times 3}$$

Square 8.

$$w=\frac{-8±\sqrt{64-4\times 3\left(-42\right)}}{2\times 3}$$

Multiply -4 times 3.

$$w=\frac{-8±\sqrt{64-12\left(-42\right)}}{2\times 3}$$

Multiply -12 times -42.

$$w=\frac{-8±\sqrt{64+504}}{2\times 3}$$

Add 64 to 504.

$$w=\frac{-8±\sqrt{568}}{2\times 3}$$

Take the square root of 568.

$$w=\frac{-8±2\sqrt{142}}{2\times 3}$$

Multiply 2 times 3.

$$w=\frac{-8±2\sqrt{142}}{6}$$

Now solve the equation w=\frac{-8±2\sqrt{142}}{6} when ± is plus. Add -8 to 2\sqrt{142}.

$$w=\frac{2\sqrt{142}-8}{6}$$

Divide -8+2\sqrt{142} by 6.

$$w=\frac{\sqrt{142}-4}{3}$$

Now solve the equation w=\frac{-8±2\sqrt{142}}{6} when ± is minus. Subtract 2\sqrt{142} from -8.

$$w=\frac{-2\sqrt{142}-8}{6}$$

Divide -8-2\sqrt{142} by 6.

$$w=\frac{-\sqrt{142}-4}{3}$$

The equation is now solved.

$$w=\frac{\sqrt{142}-4}{3} w=\frac{-\sqrt{142}-4}{3}$$

Steps for Completing the Square

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

$$3w^{2}+8w-42=0$$

Add 42 to both sides. Anything plus zero gives itself.

$$3w^{2}+8w=42$$

Divide both sides by 3.

$$\frac{3w^{2}+8w}{3}=\frac{42}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$w^{2}+\frac{8}{3}w=\frac{42}{3}$$

Divide 42 by 3.

$$w^{2}+\frac{8}{3}w=14$$

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.

$$w^{2}+\frac{8}{3}w+\left(\frac{4}{3}\right)^{2}=14+\left(\frac{4}{3}\right)^{2}$$

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

$$w^{2}+\frac{8}{3}w+\frac{16}{9}=14+\frac{16}{9}$$

Add 14 to \frac{16}{9}.

$$w^{2}+\frac{8}{3}w+\frac{16}{9}=\frac{142}{9}$$

Factor w^{2}+\frac{8}{3}w+\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}.

$$\left(w+\frac{4}{3}\right)^{2}=\frac{142}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(w+\frac{4}{3}\right)^{2}}=\sqrt{\frac{142}{9}}$$

Simplify.

$$w+\frac{4}{3}=\frac{\sqrt{142}}{3} w+\frac{4}{3}=-\frac{\sqrt{142}}{3}$$

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

$$w=\frac{\sqrt{142}-4}{3} w=\frac{-\sqrt{142}-4}{3}$$