Solve 2z^2=3-5z

$$2 z ^ { 2 } = 3 - 5 z$$

Solve for z

$$z=-3$$

$$z=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Subtract 3 from both sides.

$$2z^{2}-3=-5z$$

Add 5z to both sides.

$$2z^{2}-3+5z=0$$

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

$$2z^{2}+5z-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2z^{2}+az+bz-3. To find a and b, set up a system to be solved.

$$a+b=5 ab=2\left(-3\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 5.

$$a=-1 b=6$$

Rewrite 2z^{2}+5z-3 as \left(2z^{2}-z\right)+\left(6z-3\right).

$$\left(2z^{2}-z\right)+\left(6z-3\right)$$

Factor out z in the first and 3 in the second group.

$$z\left(2z-1\right)+3\left(2z-1\right)$$

Factor out common term 2z-1 by using distributive property.

$$\left(2z-1\right)\left(z+3\right)$$

To find equation solutions, solve 2z-1=0 and z+3=0.

$$z=\frac{1}{2} z=-3$$

Steps Using the Quadratic Formula

Subtract 3 from both sides.

$$2z^{2}-3=-5z$$

Add 5z to both sides.

$$2z^{2}-3+5z=0$$

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.

$$2z^{2}+5z-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 5 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-5±\sqrt{5^{2}-4\times 2\left(-3\right)}}{2\times 2}$$

Square 5.

$$z=\frac{-5±\sqrt{25-4\times 2\left(-3\right)}}{2\times 2}$$

Multiply -4 times 2.

$$z=\frac{-5±\sqrt{25-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$z=\frac{-5±\sqrt{25+24}}{2\times 2}$$

Add 25 to 24.

$$z=\frac{-5±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$z=\frac{-5±7}{2\times 2}$$

Multiply 2 times 2.

$$z=\frac{-5±7}{4}$$

Now solve the equation z=\frac{-5±7}{4} when ± is plus. Add -5 to 7.

$$z=\frac{2}{4}$$

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

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

Now solve the equation z=\frac{-5±7}{4} when ± is minus. Subtract 7 from -5.

$$z=-\frac{12}{4}$$

Divide -12 by 4.

$$z=-3$$

The equation is now solved.

$$z=\frac{1}{2} z=-3$$

Steps for Completing the Square

Add 5z to both sides.

$$2z^{2}+5z=3$$

Divide both sides by 2.

$$\frac{2z^{2}+5z}{2}=\frac{3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$z^{2}+\frac{5}{2}z=\frac{3}{2}$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}+\frac{5}{2}z+\left(\frac{5}{4}\right)^{2}=\frac{3}{2}+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$z^{2}+\frac{5}{2}z+\frac{25}{16}=\frac{3}{2}+\frac{25}{16}$$

Add \frac{3}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$z^{2}+\frac{5}{2}z+\frac{25}{16}=\frac{49}{16}$$

Factor z^{2}+\frac{5}{2}z+\frac{25}{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}.

$$\left(z+\frac{5}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z+\frac{5}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$z+\frac{5}{4}=\frac{7}{4} z+\frac{5}{4}=-\frac{7}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$z=\frac{1}{2} z=-3$$