Solve 2t^2+3=7t

$$2 t ^ { 2 } + 3 = 7 t$$

Solve for t

$$t=\frac{1}{2}=0.5$$

$$t=3$$

Steps Using Factoring By Grouping

Subtract 7t from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2t^{2}+at+bt+3. To find a and b, set up a system to be solved.

$$a+b=-7 ab=2\times 3=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-6 b=-1$$

Rewrite 2t^{2}-7t+3 as \left(2t^{2}-6t\right)+\left(-t+3\right).

$$\left(2t^{2}-6t\right)+\left(-t+3\right)$$

Factor out 2t in the first and -1 in the second group.

$$2t\left(t-3\right)-\left(t-3\right)$$

Factor out common term t-3 by using distributive property.

$$\left(t-3\right)\left(2t-1\right)$$

To find equation solutions, solve t-3=0 and 2t-1=0.

$$t=3 t=\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 7t from both sides.

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

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.

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

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

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

Square -7.

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

Multiply -4 times 2.

$$t=\frac{-\left(-7\right)±\sqrt{49-8\times 3}}{2\times 2}$$

Multiply -8 times 3.

$$t=\frac{-\left(-7\right)±\sqrt{49-24}}{2\times 2}$$

Add 49 to -24.

$$t=\frac{-\left(-7\right)±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$t=\frac{-\left(-7\right)±5}{2\times 2}$$

The opposite of -7 is 7.

$$t=\frac{7±5}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{7±5}{4}$$

Now solve the equation t=\frac{7±5}{4} when ± is plus. Add 7 to 5.

$$t=\frac{12}{4}$$

Divide 12 by 4.

$$t=3$$

Now solve the equation t=\frac{7±5}{4} when ± is minus. Subtract 5 from 7.

$$t=\frac{2}{4}$$

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

$$t=\frac{1}{2}$$

The equation is now solved.

$$t=3 t=\frac{1}{2}$$

Steps for Completing the Square

Subtract 7t from both sides.

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

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{7}{2}t+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}$$

Add -\frac{3}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{7}{2}t+\frac{49}{16}=\frac{25}{16}$$

Factor t^{2}-\frac{7}{2}t+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{7}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{7}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$t-\frac{7}{4}=\frac{5}{4} t-\frac{7}{4}=-\frac{5}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$t=3 t=\frac{1}{2}$$