Solve t+3+1-4t^2-2=0

$$t + 3 + 1 - 4 t ^ { 2 } - 2 = 0$$

Solve for t

$$t=\frac{\sqrt{33}+1}{8}\approx 0.843070331$$

$$t=\frac{1-\sqrt{33}}{8}\approx -0.593070331$$

Steps Using the Quadratic Formula

Add 3 and 1 to get 4.

$$t+4-4t^{2}-2=0$$

Subtract 2 from 4 to get 2.

$$t+2-4t^{2}=0$$

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.

$$-4t^{2}+t+2=0$$

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

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

Square 1.

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

Multiply -4 times -4.

$$t=\frac{-1±\sqrt{1+16\times 2}}{2\left(-4\right)}$$

Multiply 16 times 2.

$$t=\frac{-1±\sqrt{1+32}}{2\left(-4\right)}$$

Add 1 to 32.

$$t=\frac{-1±\sqrt{33}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$t=\frac{-1±\sqrt{33}}{-8}$$

Now solve the equation t=\frac{-1±\sqrt{33}}{-8} when ± is plus. Add -1 to \sqrt{33}.

$$t=\frac{\sqrt{33}-1}{-8}$$

Divide -1+\sqrt{33} by -8.

$$t=\frac{1-\sqrt{33}}{8}$$

Now solve the equation t=\frac{-1±\sqrt{33}}{-8} when ± is minus. Subtract \sqrt{33} from -1.

$$t=\frac{-\sqrt{33}-1}{-8}$$

Divide -1-\sqrt{33} by -8.

$$t=\frac{\sqrt{33}+1}{8}$$

The equation is now solved.

$$t=\frac{1-\sqrt{33}}{8} t=\frac{\sqrt{33}+1}{8}$$

Steps for Completing the Square

Add 3 and 1 to get 4.

$$t+4-4t^{2}-2=0$$

Subtract 2 from 4 to get 2.

$$t+2-4t^{2}=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$t-4t^{2}=-2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4t^{2}+t=-2$$

Divide both sides by -4.

$$\frac{-4t^{2}+t}{-4}=-\frac{2}{-4}$$

Dividing by -4 undoes the multiplication by -4.

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

Divide 1 by -4.

$$t^{2}-\frac{1}{4}t=-\frac{2}{-4}$$

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out 2.

$$t^{2}-\frac{1}{4}t=\frac{1}{2}$$

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{1}{4}t+\left(-\frac{1}{8}\right)^{2}=\frac{1}{2}+\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{1}{4}t+\frac{1}{64}=\frac{1}{2}+\frac{1}{64}$$

Add \frac{1}{2} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{1}{4}t+\frac{1}{64}=\frac{33}{64}$$

Factor t^{2}-\frac{1}{4}t+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{1}{8}\right)^{2}=\frac{33}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{1}{8}\right)^{2}}=\sqrt{\frac{33}{64}}$$

Simplify.

$$t-\frac{1}{8}=\frac{\sqrt{33}}{8} t-\frac{1}{8}=-\frac{\sqrt{33}}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$t=\frac{\sqrt{33}+1}{8} t=\frac{1-\sqrt{33}}{8}$$