Solve 5+23z=10z^2

$$5 + 23 z = 10 z ^ { 2 }$$

Solve for z

$$z=-\frac{1}{5}=-0.2$$

$$z = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

Subtract 10z^{2} from both sides.

$$5+23z-10z^{2}=0$$

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

$$-10z^{2}+23z+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -10z^{2}+az+bz+5. To find a and b, set up a system to be solved.

$$a+b=23 ab=-10\times 5=-50$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -50.

$$-1,50 -2,25 -5,10$$

Calculate the sum for each pair.

$$-1+50=49 -2+25=23 -5+10=5$$

The solution is the pair that gives sum 23.

$$a=25 b=-2$$

Rewrite -10z^{2}+23z+5 as \left(-10z^{2}+25z\right)+\left(-2z+5\right).

$$\left(-10z^{2}+25z\right)+\left(-2z+5\right)$$

Factor out -5z in the first and -1 in the second group.

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

Factor out common term 2z-5 by using distributive property.

$$\left(2z-5\right)\left(-5z-1\right)$$

To find equation solutions, solve 2z-5=0 and -5z-1=0.

$$z=\frac{5}{2} z=-\frac{1}{5}$$

Steps Using the Quadratic Formula

Subtract 10z^{2} from both sides.

$$5+23z-10z^{2}=0$$

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.

$$-10z^{2}+23z+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 23 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-23±\sqrt{23^{2}-4\left(-10\right)\times 5}}{2\left(-10\right)}$$

Square 23.

$$z=\frac{-23±\sqrt{529-4\left(-10\right)\times 5}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$z=\frac{-23±\sqrt{529+40\times 5}}{2\left(-10\right)}$$

Multiply 40 times 5.

$$z=\frac{-23±\sqrt{529+200}}{2\left(-10\right)}$$

Add 529 to 200.

$$z=\frac{-23±\sqrt{729}}{2\left(-10\right)}$$

Take the square root of 729.

$$z=\frac{-23±27}{2\left(-10\right)}$$

Multiply 2 times -10.

$$z=\frac{-23±27}{-20}$$

Now solve the equation z=\frac{-23±27}{-20} when ± is plus. Add -23 to 27.

$$z=\frac{4}{-20}$$

Reduce the fraction \frac{4}{-20} to lowest terms by extracting and canceling out 4.

$$z=-\frac{1}{5}$$

Now solve the equation z=\frac{-23±27}{-20} when ± is minus. Subtract 27 from -23.

$$z=-\frac{50}{-20}$$

Reduce the fraction \frac{-50}{-20} to lowest terms by extracting and canceling out 10.

$$z=\frac{5}{2}$$

The equation is now solved.

$$z=-\frac{1}{5} z=\frac{5}{2}$$

Steps for Completing the Square

Subtract 10z^{2} from both sides.

$$5+23z-10z^{2}=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$23z-10z^{2}=-5$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-10z^{2}+23z=-5$$

Divide both sides by -10.

$$\frac{-10z^{2}+23z}{-10}=-\frac{5}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$z^{2}+\frac{23}{-10}z=-\frac{5}{-10}$$

Divide 23 by -10.

$$z^{2}-\frac{23}{10}z=-\frac{5}{-10}$$

Reduce the fraction \frac{-5}{-10} to lowest terms by extracting and canceling out 5.

$$z^{2}-\frac{23}{10}z=\frac{1}{2}$$

Divide -\frac{23}{10}, the coefficient of the x term, by 2 to get -\frac{23}{20}. Then add the square of -\frac{23}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-\frac{23}{10}z+\left(-\frac{23}{20}\right)^{2}=\frac{1}{2}+\left(-\frac{23}{20}\right)^{2}$$

Square -\frac{23}{20} by squaring both the numerator and the denominator of the fraction.

$$z^{2}-\frac{23}{10}z+\frac{529}{400}=\frac{1}{2}+\frac{529}{400}$$

Add \frac{1}{2} to \frac{529}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$z^{2}-\frac{23}{10}z+\frac{529}{400}=\frac{729}{400}$$

Factor z^{2}-\frac{23}{10}z+\frac{529}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(z-\frac{23}{20}\right)^{2}=\frac{729}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{23}{20}\right)^{2}}=\sqrt{\frac{729}{400}}$$

Simplify.

$$z-\frac{23}{20}=\frac{27}{20} z-\frac{23}{20}=-\frac{27}{20}$$

Add \frac{23}{20} to both sides of the equation.

$$z=\frac{5}{2} z=-\frac{1}{5}$$