Solve f(x)=3x^2-5x-2quad0quadg(x)=2x-7

$$f ( x ) = 3 x ^ { 2 } - 5 x - 2 \quad 0 \quad g ( x ) = 2 x - 7$$

Solve for x (complex solution)

$$x=\frac{7+\sqrt{35}i}{6}$$

$$x=\frac{-\sqrt{35}i+7}{6}$$

Steps Using the Quadratic Formula

Multiply 2 and 0 to get 0.

$$3x^{2}-5x-0gx=2x-7$$

Anything times zero gives zero.

$$3x^{2}-5x-0=2x-7$$

Subtract 2x from both sides.

$$3x^{2}-5x-0-2x=-7$$

Add 7 to both sides.

$$3x^{2}-5x-0-2x+7=0$$

Reorder the terms.

$$3x^{2}-5x-2x+7=0$$

Combine -5x and -2x to get -7x.

$$3x^{2}-7x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -7 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\times 7}}{2\times 3}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 3\times 7}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-7\right)±\sqrt{49-12\times 7}}{2\times 3}$$

Multiply -12 times 7.

$$x=\frac{-\left(-7\right)±\sqrt{49-84}}{2\times 3}$$

Add 49 to -84.

$$x=\frac{-\left(-7\right)±\sqrt{-35}}{2\times 3}$$

Take the square root of -35.

$$x=\frac{-\left(-7\right)±\sqrt{35}i}{2\times 3}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{35}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{7±\sqrt{35}i}{6}$$

Now solve the equation x=\frac{7±\sqrt{35}i}{6} when ± is plus. Add 7 to i\sqrt{35}.

$$x=\frac{7+\sqrt{35}i}{6}$$

Now solve the equation x=\frac{7±\sqrt{35}i}{6} when ± is minus. Subtract i\sqrt{35} from 7.

$$x=\frac{-\sqrt{35}i+7}{6}$$

The equation is now solved.

$$x=\frac{7+\sqrt{35}i}{6} x=\frac{-\sqrt{35}i+7}{6}$$

Steps for Completing the Square

Multiply 2 and 0 to get 0.

$$3x^{2}-5x-0gx=2x-7$$

Anything times zero gives zero.

$$3x^{2}-5x-0=2x-7$$

Subtract 2x from both sides.

$$3x^{2}-5x-0-2x=-7$$

Reorder the terms.

$$3x^{2}-5x-2x=-7$$

Combine -5x and -2x to get -7x.

$$3x^{2}-7x=-7$$

Divide both sides by 3.

$$\frac{3x^{2}-7x}{3}=-\frac{7}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{7}{3}x=-\frac{7}{3}$$

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{3}x+\left(-\frac{7}{6}\right)^{2}=-\frac{7}{3}+\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=-\frac{7}{3}+\frac{49}{36}$$

Add -\frac{7}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=-\frac{35}{36}$$

Factor x^{2}-\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{6}\right)^{2}=-\frac{35}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{6}\right)^{2}}=\sqrt{-\frac{35}{36}}$$

Simplify.

$$x-\frac{7}{6}=\frac{\sqrt{35}i}{6} x-\frac{7}{6}=-\frac{\sqrt{35}i}{6}$$

Add \frac{7}{6} to both sides of the equation.

$$x=\frac{7+\sqrt{35}i}{6} x=\frac{-\sqrt{35}i+7}{6}$$

Solve for g (complex solution)

$$g\in \mathrm{C}$$

$$x=\frac{7+\sqrt{35}i}{6}\text{ or }x=\frac{-\sqrt{35}i+7}{6}$$