Solve (2z-3)^2=11z-19

$$( 2 z - 3 ) ^ { 2 } = 11 z - 19$$

Solve for z

$$z=4$$

$$z = \frac{7}{4} = 1\frac{3}{4} = 1.75$$

Steps Using Factoring By Grouping

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

$$4z^{2}-12z+9=11z-19$$

Subtract 11z from both sides.

$$4z^{2}-12z+9-11z=-19$$

Combine -12z and -11z to get -23z.

$$4z^{2}-23z+9=-19$$

Add 19 to both sides.

$$4z^{2}-23z+9+19=0$$

Add 9 and 19 to get 28.

$$4z^{2}-23z+28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4z^{2}+az+bz+28. To find a and b, set up a system to be solved.

$$a+b=-23 ab=4\times 28=112$$

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 112.

$$-1,-112 -2,-56 -4,-28 -7,-16 -8,-14$$

Calculate the sum for each pair.

$$-1-112=-113 -2-56=-58 -4-28=-32 -7-16=-23 -8-14=-22$$

The solution is the pair that gives sum -23.

$$a=-16 b=-7$$

Rewrite 4z^{2}-23z+28 as \left(4z^{2}-16z\right)+\left(-7z+28\right).

$$\left(4z^{2}-16z\right)+\left(-7z+28\right)$$

Factor out 4z in the first and -7 in the second group.

$$4z\left(z-4\right)-7\left(z-4\right)$$

Factor out common term z-4 by using distributive property.

$$\left(z-4\right)\left(4z-7\right)$$

To find equation solutions, solve z-4=0 and 4z-7=0.

$$z=4 z=\frac{7}{4}$$

Steps Using the Quadratic Formula

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

$$4z^{2}-12z+9=11z-19$$

Subtract 11z from both sides.

$$4z^{2}-12z+9-11z=-19$$

Combine -12z and -11z to get -23z.

$$4z^{2}-23z+9=-19$$

Add 19 to both sides.

$$4z^{2}-23z+9+19=0$$

Add 9 and 19 to get 28.

$$4z^{2}-23z+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -23 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 4\times 28}}{2\times 4}$$

Square -23.

$$z=\frac{-\left(-23\right)±\sqrt{529-4\times 4\times 28}}{2\times 4}$$

Multiply -4 times 4.

$$z=\frac{-\left(-23\right)±\sqrt{529-16\times 28}}{2\times 4}$$

Multiply -16 times 28.

$$z=\frac{-\left(-23\right)±\sqrt{529-448}}{2\times 4}$$

Add 529 to -448.

$$z=\frac{-\left(-23\right)±\sqrt{81}}{2\times 4}$$

Take the square root of 81.

$$z=\frac{-\left(-23\right)±9}{2\times 4}$$

The opposite of -23 is 23.

$$z=\frac{23±9}{2\times 4}$$

Multiply 2 times 4.

$$z=\frac{23±9}{8}$$

Now solve the equation z=\frac{23±9}{8} when ± is plus. Add 23 to 9.

$$z=\frac{32}{8}$$

Divide 32 by 8.

$$z=4$$

Now solve the equation z=\frac{23±9}{8} when ± is minus. Subtract 9 from 23.

$$z=\frac{14}{8}$$

Reduce the fraction \frac{14}{8} to lowest terms by extracting and canceling out 2.

$$z=\frac{7}{4}$$

The equation is now solved.

$$z=4 z=\frac{7}{4}$$

Steps for Completing the Square

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

$$4z^{2}-12z+9=11z-19$$

Subtract 11z from both sides.

$$4z^{2}-12z+9-11z=-19$$

Combine -12z and -11z to get -23z.

$$4z^{2}-23z+9=-19$$

Subtract 9 from both sides.

$$4z^{2}-23z=-19-9$$

Subtract 9 from -19 to get -28.

$$4z^{2}-23z=-28$$

Divide both sides by 4.

$$\frac{4z^{2}-23z}{4}=-\frac{28}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$z^{2}-\frac{23}{4}z=-\frac{28}{4}$$

Divide -28 by 4.

$$z^{2}-\frac{23}{4}z=-7$$

Divide -\frac{23}{4}, the coefficient of the x term, by 2 to get -\frac{23}{8}. Then add the square of -\frac{23}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-\frac{23}{4}z+\left(-\frac{23}{8}\right)^{2}=-7+\left(-\frac{23}{8}\right)^{2}$$

Square -\frac{23}{8} by squaring both the numerator and the denominator of the fraction.

$$z^{2}-\frac{23}{4}z+\frac{529}{64}=-7+\frac{529}{64}$$

Add -7 to \frac{529}{64}.

$$z^{2}-\frac{23}{4}z+\frac{529}{64}=\frac{81}{64}$$

Factor z^{2}-\frac{23}{4}z+\frac{529}{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}.

$$\left(z-\frac{23}{8}\right)^{2}=\frac{81}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{23}{8}\right)^{2}}=\sqrt{\frac{81}{64}}$$

Simplify.

$$z-\frac{23}{8}=\frac{9}{8} z-\frac{23}{8}=-\frac{9}{8}$$

Add \frac{23}{8} to both sides of the equation.

$$z=4 z=\frac{7}{4}$$