Solve t^2+11+t-42=0

$${ t }^{ 2 } +11+t-42 = 0$$

Solve for t

$$t = \frac{5 \sqrt{5} - 1}{2} \approx 5.090169944$$

$$t=\frac{-5\sqrt{5}-1}{2}\approx -6.090169944$$

Steps Using the Quadratic Formula

Subtract 42 from 11 to get -31.

$$t^{2}-31+t=0$$

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.

$$t^{2}+t-31=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -31 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-1±\sqrt{1^{2}-4\left(-31\right)}}{2}$$

Square 1.

$$t=\frac{-1±\sqrt{1-4\left(-31\right)}}{2}$$

Multiply -4 times -31.

$$t=\frac{-1±\sqrt{1+124}}{2}$$

Add 1 to 124.

$$t=\frac{-1±\sqrt{125}}{2}$$

Take the square root of 125.

$$t=\frac{-1±5\sqrt{5}}{2}$$

Now solve the equation t=\frac{-1±5\sqrt{5}}{2} when ± is plus. Add -1 to 5\sqrt{5}.

$$t=\frac{5\sqrt{5}-1}{2}$$

Now solve the equation t=\frac{-1±5\sqrt{5}}{2} when ± is minus. Subtract 5\sqrt{5} from -1.

$$t=\frac{-5\sqrt{5}-1}{2}$$

The equation is now solved.

$$t=\frac{5\sqrt{5}-1}{2} t=\frac{-5\sqrt{5}-1}{2}$$

Steps for Completing the Square

Subtract 42 from 11 to get -31.

$$t^{2}-31+t=0$$

Add 31 to both sides. Anything plus zero gives itself.

$$t^{2}+t=31$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+t+\left(\frac{1}{2}\right)^{2}=31+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+t+\frac{1}{4}=31+\frac{1}{4}$$

Add 31 to \frac{1}{4}.

$$t^{2}+t+\frac{1}{4}=\frac{125}{4}$$

Factor t^{2}+t+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{1}{2}\right)^{2}=\frac{125}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{1}{2}\right)^{2}}=\sqrt{\frac{125}{4}}$$

Simplify.

$$t+\frac{1}{2}=\frac{5\sqrt{5}}{2} t+\frac{1}{2}=-\frac{5\sqrt{5}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$t=\frac{5\sqrt{5}-1}{2} t=\frac{-5\sqrt{5}-1}{2}$$