Solve 2z^2-30=11z

$$2 z ^ { 2 } - 30 = 11 z$$

Solve for z

$$z=-2$$

$$z = \frac{15}{2} = 7\frac{1}{2} = 7.5$$

Steps Using Factoring By Grouping

Subtract 11z from both sides.

$$2z^{2}-30-11z=0$$

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

$$2z^{2}-11z-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2z^{2}+az+bz-30. To find a and b, set up a system to be solved.

$$a+b=-11 ab=2\left(-30\right)=-60$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -11.

$$a=-15 b=4$$

Rewrite 2z^{2}-11z-30 as \left(2z^{2}-15z\right)+\left(4z-30\right).

$$\left(2z^{2}-15z\right)+\left(4z-30\right)$$

Factor out z in the first and 2 in the second group.

$$z\left(2z-15\right)+2\left(2z-15\right)$$

Factor out common term 2z-15 by using distributive property.

$$\left(2z-15\right)\left(z+2\right)$$

To find equation solutions, solve 2z-15=0 and z+2=0.

$$z=\frac{15}{2} z=-2$$

Steps Using the Quadratic Formula

Subtract 11z from both sides.

$$2z^{2}-30-11z=0$$

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.

$$2z^{2}-11z-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -11 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\left(-30\right)}}{2\times 2}$$

Square -11.

$$z=\frac{-\left(-11\right)±\sqrt{121-4\times 2\left(-30\right)}}{2\times 2}$$

Multiply -4 times 2.

$$z=\frac{-\left(-11\right)±\sqrt{121-8\left(-30\right)}}{2\times 2}$$

Multiply -8 times -30.

$$z=\frac{-\left(-11\right)±\sqrt{121+240}}{2\times 2}$$

Add 121 to 240.

$$z=\frac{-\left(-11\right)±\sqrt{361}}{2\times 2}$$

Take the square root of 361.

$$z=\frac{-\left(-11\right)±19}{2\times 2}$$

The opposite of -11 is 11.

$$z=\frac{11±19}{2\times 2}$$

Multiply 2 times 2.

$$z=\frac{11±19}{4}$$

Now solve the equation z=\frac{11±19}{4} when ± is plus. Add 11 to 19.

$$z=\frac{30}{4}$$

Reduce the fraction \frac{30}{4} to lowest terms by extracting and canceling out 2.

$$z=\frac{15}{2}$$

Now solve the equation z=\frac{11±19}{4} when ± is minus. Subtract 19 from 11.

$$z=-\frac{8}{4}$$

Divide -8 by 4.

$$z=-2$$

The equation is now solved.

$$z=\frac{15}{2} z=-2$$

Steps for Completing the Square

Subtract 11z from both sides.

$$2z^{2}-30-11z=0$$

Add 30 to both sides. Anything plus zero gives itself.

$$2z^{2}-11z=30$$

Divide both sides by 2.

$$\frac{2z^{2}-11z}{2}=\frac{30}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$z^{2}-\frac{11}{2}z=\frac{30}{2}$$

Divide 30 by 2.

$$z^{2}-\frac{11}{2}z=15$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-\frac{11}{2}z+\left(-\frac{11}{4}\right)^{2}=15+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$z^{2}-\frac{11}{2}z+\frac{121}{16}=15+\frac{121}{16}$$

Add 15 to \frac{121}{16}.

$$z^{2}-\frac{11}{2}z+\frac{121}{16}=\frac{361}{16}$$

Factor z^{2}-\frac{11}{2}z+\frac{121}{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}.

$$\left(z-\frac{11}{4}\right)^{2}=\frac{361}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{11}{4}\right)^{2}}=\sqrt{\frac{361}{16}}$$

Simplify.

$$z-\frac{11}{4}=\frac{19}{4} z-\frac{11}{4}=-\frac{19}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$z=\frac{15}{2} z=-2$$