Solve (2t-3)^2-8(2t-3)+7=0

$$( 2 t - 3 ) ^ { 2 } - 8 ( 2 t - 3 ) + 7 = 0$$

Solve for t

$$t=2$$

$$t=5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2t-3\right)^{2}.

$$4t^{2}-12t+9-8\left(2t-3\right)+7=0$$

Use the distributive property to multiply -8 by 2t-3.

$$4t^{2}-12t+9-16t+24+7=0$$

Combine -12t and -16t to get -28t.

$$4t^{2}-28t+9+24+7=0$$

Add 9 and 24 to get 33.

$$4t^{2}-28t+33+7=0$$

Add 33 and 7 to get 40.

$$4t^{2}-28t+40=0$$

Divide both sides by 4.

$$t^{2}-7t+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt+10. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 10=10$$

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 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -7.

$$a=-5 b=-2$$

Rewrite t^{2}-7t+10 as \left(t^{2}-5t\right)+\left(-2t+10\right).

$$\left(t^{2}-5t\right)+\left(-2t+10\right)$$

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

$$t\left(t-5\right)-2\left(t-5\right)$$

Factor out common term t-5 by using distributive property.

$$\left(t-5\right)\left(t-2\right)$$

To find equation solutions, solve t-5=0 and t-2=0.

$$t=5 t=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2t-3\right)^{2}.

$$4t^{2}-12t+9-8\left(2t-3\right)+7=0$$

Use the distributive property to multiply -8 by 2t-3.

$$4t^{2}-12t+9-16t+24+7=0$$

Combine -12t and -16t to get -28t.

$$4t^{2}-28t+9+24+7=0$$

Add 9 and 24 to get 33.

$$4t^{2}-28t+33+7=0$$

Add 33 and 7 to get 40.

$$4t^{2}-28t+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -28 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 4\times 40}}{2\times 4}$$

Square -28.

$$t=\frac{-\left(-28\right)±\sqrt{784-4\times 4\times 40}}{2\times 4}$$

Multiply -4 times 4.

$$t=\frac{-\left(-28\right)±\sqrt{784-16\times 40}}{2\times 4}$$

Multiply -16 times 40.

$$t=\frac{-\left(-28\right)±\sqrt{784-640}}{2\times 4}$$

Add 784 to -640.

$$t=\frac{-\left(-28\right)±\sqrt{144}}{2\times 4}$$

Take the square root of 144.

$$t=\frac{-\left(-28\right)±12}{2\times 4}$$

The opposite of -28 is 28.

$$t=\frac{28±12}{2\times 4}$$

Multiply 2 times 4.

$$t=\frac{28±12}{8}$$

Now solve the equation t=\frac{28±12}{8} when ± is plus. Add 28 to 12.

$$t=\frac{40}{8}$$

Divide 40 by 8.

$$t=5$$

Now solve the equation t=\frac{28±12}{8} when ± is minus. Subtract 12 from 28.

$$t=\frac{16}{8}$$

Divide 16 by 8.

$$t=2$$

The equation is now solved.

$$t=5 t=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2t-3\right)^{2}.

$$4t^{2}-12t+9-8\left(2t-3\right)+7=0$$

Use the distributive property to multiply -8 by 2t-3.

$$4t^{2}-12t+9-16t+24+7=0$$

Combine -12t and -16t to get -28t.

$$4t^{2}-28t+9+24+7=0$$

Add 9 and 24 to get 33.

$$4t^{2}-28t+33+7=0$$

Add 33 and 7 to get 40.

$$4t^{2}-28t+40=0$$

Subtract 40 from both sides. Anything subtracted from zero gives its negation.

$$4t^{2}-28t=-40$$

Divide both sides by 4.

$$\frac{4t^{2}-28t}{4}=-\frac{40}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$t^{2}+\left(-\frac{28}{4}\right)t=-\frac{40}{4}$$

Divide -28 by 4.

$$t^{2}-7t=-\frac{40}{4}$$

Divide -40 by 4.

$$t^{2}-7t=-10$$

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

$$t^{2}-7t+\left(-\frac{7}{2}\right)^{2}=-10+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-7t+\frac{49}{4}=-10+\frac{49}{4}$$

Add -10 to \frac{49}{4}.

$$t^{2}-7t+\frac{49}{4}=\frac{9}{4}$$

Factor t^{2}-7t+\frac{49}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{7}{2} to both sides of the equation.

$$t=5 t=2$$