Solve (4t-7)^2+9=25

$$( 4 t - 7 ) ^ { 2 } + 9 = 25$$

Solve for t

$$t = \frac{11}{4} = 2\frac{3}{4} = 2.75$$

$$t=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4t-7\right)^{2}.

$$16t^{2}-56t+49+9=25$$

Add 49 and 9 to get 58.

$$16t^{2}-56t+58=25$$

Subtract 25 from both sides.

$$16t^{2}-56t+58-25=0$$

Subtract 25 from 58 to get 33.

$$16t^{2}-56t+33=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16t^{2}+at+bt+33. To find a and b, set up a system to be solved.

$$a+b=-56 ab=16\times 33=528$$

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 528.

$$-1,-528 -2,-264 -3,-176 -4,-132 -6,-88 -8,-66 -11,-48 -12,-44 -16,-33 -22,-24$$

Calculate the sum for each pair.

$$-1-528=-529 -2-264=-266 -3-176=-179 -4-132=-136 -6-88=-94 -8-66=-74 -11-48=-59 -12-44=-56 -16-33=-49 -22-24=-46$$

The solution is the pair that gives sum -56.

$$a=-44 b=-12$$

Rewrite 16t^{2}-56t+33 as \left(16t^{2}-44t\right)+\left(-12t+33\right).

$$\left(16t^{2}-44t\right)+\left(-12t+33\right)$$

Factor out 4t in the first and -3 in the second group.

$$4t\left(4t-11\right)-3\left(4t-11\right)$$

Factor out common term 4t-11 by using distributive property.

$$\left(4t-11\right)\left(4t-3\right)$$

To find equation solutions, solve 4t-11=0 and 4t-3=0.

$$t=\frac{11}{4} t=\frac{3}{4}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4t-7\right)^{2}.

$$16t^{2}-56t+49+9=25$$

Add 49 and 9 to get 58.

$$16t^{2}-56t+58=25$$

Subtract 25 from both sides.

$$16t^{2}-56t+58-25=0$$

Subtract 25 from 58 to get 33.

$$16t^{2}-56t+33=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -56 for b, and 33 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-56\right)±\sqrt{\left(-56\right)^{2}-4\times 16\times 33}}{2\times 16}$$

Square -56.

$$t=\frac{-\left(-56\right)±\sqrt{3136-4\times 16\times 33}}{2\times 16}$$

Multiply -4 times 16.

$$t=\frac{-\left(-56\right)±\sqrt{3136-64\times 33}}{2\times 16}$$

Multiply -64 times 33.

$$t=\frac{-\left(-56\right)±\sqrt{3136-2112}}{2\times 16}$$

Add 3136 to -2112.

$$t=\frac{-\left(-56\right)±\sqrt{1024}}{2\times 16}$$

Take the square root of 1024.

$$t=\frac{-\left(-56\right)±32}{2\times 16}$$

The opposite of -56 is 56.

$$t=\frac{56±32}{2\times 16}$$

Multiply 2 times 16.

$$t=\frac{56±32}{32}$$

Now solve the equation t=\frac{56±32}{32} when ± is plus. Add 56 to 32.

$$t=\frac{88}{32}$$

Reduce the fraction \frac{88}{32} to lowest terms by extracting and canceling out 8.

$$t=\frac{11}{4}$$

Now solve the equation t=\frac{56±32}{32} when ± is minus. Subtract 32 from 56.

$$t=\frac{24}{32}$$

Reduce the fraction \frac{24}{32} to lowest terms by extracting and canceling out 8.

$$t=\frac{3}{4}$$

The equation is now solved.

$$t=\frac{11}{4} t=\frac{3}{4}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4t-7\right)^{2}.

$$16t^{2}-56t+49+9=25$$

Add 49 and 9 to get 58.

$$16t^{2}-56t+58=25$$

Subtract 58 from both sides.

$$16t^{2}-56t=25-58$$

Subtract 58 from 25 to get -33.

$$16t^{2}-56t=-33$$

Divide both sides by 16.

$$\frac{16t^{2}-56t}{16}=-\frac{33}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$t^{2}+\left(-\frac{56}{16}\right)t=-\frac{33}{16}$$

Reduce the fraction \frac{-56}{16} to lowest terms by extracting and canceling out 8.

$$t^{2}-\frac{7}{2}t=-\frac{33}{16}$$

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.

$$t^{2}-\frac{7}{2}t+\left(-\frac{7}{4}\right)^{2}=-\frac{33}{16}+\left(-\frac{7}{4}\right)^{2}$$

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

$$t^{2}-\frac{7}{2}t+\frac{49}{16}=\frac{-33+49}{16}$$

Add -\frac{33}{16} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{7}{2}t+\frac{49}{16}=1$$

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}.

$$\left(t-\frac{7}{4}\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{7}{4}\right)^{2}}=\sqrt{1}$$

Simplify.

$$t-\frac{7}{4}=1 t-\frac{7}{4}=-1$$

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

$$t=\frac{11}{4} t=\frac{3}{4}$$