Solve 9u^2+3=42u-50

$$9 u ^ { 2 } + 3 = 42 u - 50$$

Solve for u

$$u=\frac{7}{3}+\frac{2}{3}i\approx 2.333333333+0.666666667i$$

$$u=\frac{7}{3}-\frac{2}{3}i\approx 2.333333333-0.666666667i$$

Steps Using the Quadratic Formula

Subtract 42u from both sides.

$$9u^{2}+3-42u=-50$$

Add 50 to both sides.

$$9u^{2}+3-42u+50=0$$

Add 3 and 50 to get 53.

$$9u^{2}+53-42u=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$9u^{2}-42u+53=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -42 for b, and 53 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$u=\frac{-\left(-42\right)±\sqrt{\left(-42\right)^{2}-4\times 9\times 53}}{2\times 9}$$

Square -42.

$$u=\frac{-\left(-42\right)±\sqrt{1764-4\times 9\times 53}}{2\times 9}$$

Multiply -4 times 9.

$$u=\frac{-\left(-42\right)±\sqrt{1764-36\times 53}}{2\times 9}$$

Multiply -36 times 53.

$$u=\frac{-\left(-42\right)±\sqrt{1764-1908}}{2\times 9}$$

Add 1764 to -1908.

$$u=\frac{-\left(-42\right)±\sqrt{-144}}{2\times 9}$$

Take the square root of -144.

$$u=\frac{-\left(-42\right)±12i}{2\times 9}$$

The opposite of -42 is 42.

$$u=\frac{42±12i}{2\times 9}$$

Multiply 2 times 9.

$$u=\frac{42±12i}{18}$$

Now solve the equation u=\frac{42±12i}{18} when ± is plus. Add 42 to 12i.

$$u=\frac{42+12i}{18}$$

Divide 42+12i by 18.

$$u=\frac{7}{3}+\frac{2}{3}i$$

Now solve the equation u=\frac{42±12i}{18} when ± is minus. Subtract 12i from 42.

$$u=\frac{42-12i}{18}$$

Divide 42-12i by 18.

$$u=\frac{7}{3}-\frac{2}{3}i$$

The equation is now solved.

$$u=\frac{7}{3}+\frac{2}{3}i u=\frac{7}{3}-\frac{2}{3}i$$

Steps for Completing the Square

Subtract 42u from both sides.

$$9u^{2}+3-42u=-50$$

Subtract 3 from both sides.

$$9u^{2}-42u=-50-3$$

Subtract 3 from -50 to get -53.

$$9u^{2}-42u=-53$$

Divide both sides by 9.

$$\frac{9u^{2}-42u}{9}=-\frac{53}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$u^{2}+\left(-\frac{42}{9}\right)u=-\frac{53}{9}$$

Reduce the fraction \frac{-42}{9} to lowest terms by extracting and canceling out 3.

$$u^{2}-\frac{14}{3}u=-\frac{53}{9}$$

Divide -\frac{14}{3}, the coefficient of the x term, by 2 to get -\frac{7}{3}. Then add the square of -\frac{7}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$u^{2}-\frac{14}{3}u+\left(-\frac{7}{3}\right)^{2}=-\frac{53}{9}+\left(-\frac{7}{3}\right)^{2}$$

Square -\frac{7}{3} by squaring both the numerator and the denominator of the fraction.

$$u^{2}-\frac{14}{3}u+\frac{49}{9}=\frac{-53+49}{9}$$

Add -\frac{53}{9} to \frac{49}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$u^{2}-\frac{14}{3}u+\frac{49}{9}=-\frac{4}{9}$$

Factor u^{2}-\frac{14}{3}u+\frac{49}{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}.

$$\left(u-\frac{7}{3}\right)^{2}=-\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(u-\frac{7}{3}\right)^{2}}=\sqrt{-\frac{4}{9}}$$

Simplify.

$$u-\frac{7}{3}=\frac{2}{3}i u-\frac{7}{3}=-\frac{2}{3}i$$

Add \frac{7}{3} to both sides of the equation.

$$u=\frac{7}{3}+\frac{2}{3}i u=\frac{7}{3}-\frac{2}{3}i$$