Solve 6z^2+28z+50=5z^2+46z-30

$$6 z ^ { 2 } + 28 z + 50 = 5 z ^ { 2 } + 46 z - 30$$

Solve for z

$$z=8$$

$$z=10$$

Steps Using Factoring

Subtract 5z^{2} from both sides.

$$6z^{2}+28z+50-5z^{2}=46z-30$$

Combine 6z^{2} and -5z^{2} to get z^{2}.

$$z^{2}+28z+50=46z-30$$

Subtract 46z from both sides.

$$z^{2}+28z+50-46z=-30$$

Combine 28z and -46z to get -18z.

$$z^{2}-18z+50=-30$$

Add 30 to both sides.

$$z^{2}-18z+50+30=0$$

Add 50 and 30 to get 80.

$$z^{2}-18z+80=0$$

To solve the equation, factor z^{2}-18z+80 using formula z^{2}+\left(a+b\right)z+ab=\left(z+a\right)\left(z+b\right). To find a and b, set up a system to be solved.

$$a+b=-18 ab=80$$

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 80.

$$-1,-80 -2,-40 -4,-20 -5,-16 -8,-10$$

Calculate the sum for each pair.

$$-1-80=-81 -2-40=-42 -4-20=-24 -5-16=-21 -8-10=-18$$

The solution is the pair that gives sum -18.

$$a=-10 b=-8$$

Rewrite factored expression \left(z+a\right)\left(z+b\right) using the obtained values.

$$\left(z-10\right)\left(z-8\right)$$

To find equation solutions, solve z-10=0 and z-8=0.

$$z=10 z=8$$

Steps Using Factoring By Grouping

Subtract 5z^{2} from both sides.

$$6z^{2}+28z+50-5z^{2}=46z-30$$

Combine 6z^{2} and -5z^{2} to get z^{2}.

$$z^{2}+28z+50=46z-30$$

Subtract 46z from both sides.

$$z^{2}+28z+50-46z=-30$$

Combine 28z and -46z to get -18z.

$$z^{2}-18z+50=-30$$

Add 30 to both sides.

$$z^{2}-18z+50+30=0$$

Add 50 and 30 to get 80.

$$z^{2}-18z+80=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as z^{2}+az+bz+80. To find a and b, set up a system to be solved.

$$a+b=-18 ab=1\times 80=80$$

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 80.

$$-1,-80 -2,-40 -4,-20 -5,-16 -8,-10$$

Calculate the sum for each pair.

$$-1-80=-81 -2-40=-42 -4-20=-24 -5-16=-21 -8-10=-18$$

The solution is the pair that gives sum -18.

$$a=-10 b=-8$$

Rewrite z^{2}-18z+80 as \left(z^{2}-10z\right)+\left(-8z+80\right).

$$\left(z^{2}-10z\right)+\left(-8z+80\right)$$

Factor out z in the first and -8 in the second group.

$$z\left(z-10\right)-8\left(z-10\right)$$

Factor out common term z-10 by using distributive property.

$$\left(z-10\right)\left(z-8\right)$$

To find equation solutions, solve z-10=0 and z-8=0.

$$z=10 z=8$$

Steps Using the Quadratic Formula

Subtract 5z^{2} from both sides.

$$6z^{2}+28z+50-5z^{2}=46z-30$$

Combine 6z^{2} and -5z^{2} to get z^{2}.

$$z^{2}+28z+50=46z-30$$

Subtract 46z from both sides.

$$z^{2}+28z+50-46z=-30$$

Combine 28z and -46z to get -18z.

$$z^{2}-18z+50=-30$$

Add 30 to both sides.

$$z^{2}-18z+50+30=0$$

Add 50 and 30 to get 80.

$$z^{2}-18z+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -18 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 80}}{2}$$

Square -18.

$$z=\frac{-\left(-18\right)±\sqrt{324-4\times 80}}{2}$$

Multiply -4 times 80.

$$z=\frac{-\left(-18\right)±\sqrt{324-320}}{2}$$

Add 324 to -320.

$$z=\frac{-\left(-18\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$z=\frac{-\left(-18\right)±2}{2}$$

The opposite of -18 is 18.

$$z=\frac{18±2}{2}$$

Now solve the equation z=\frac{18±2}{2} when ± is plus. Add 18 to 2.

$$z=\frac{20}{2}$$

Divide 20 by 2.

$$z=10$$

Now solve the equation z=\frac{18±2}{2} when ± is minus. Subtract 2 from 18.

$$z=\frac{16}{2}$$

Divide 16 by 2.

$$z=8$$

The equation is now solved.

$$z=10 z=8$$

Steps for Completing the Square

Subtract 5z^{2} from both sides.

$$6z^{2}+28z+50-5z^{2}=46z-30$$

Combine 6z^{2} and -5z^{2} to get z^{2}.

$$z^{2}+28z+50=46z-30$$

Subtract 46z from both sides.

$$z^{2}+28z+50-46z=-30$$

Combine 28z and -46z to get -18z.

$$z^{2}-18z+50=-30$$

Subtract 50 from both sides.

$$z^{2}-18z=-30-50$$

Subtract 50 from -30 to get -80.

$$z^{2}-18z=-80$$

Divide -18, the coefficient of the x term, by 2 to get -9. Then add the square of -9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-18z+\left(-9\right)^{2}=-80+\left(-9\right)^{2}$$

Square -9.

$$z^{2}-18z+81=-80+81$$

Add -80 to 81.

$$z^{2}-18z+81=1$$

Factor z^{2}-18z+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(z-9\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-9\right)^{2}}=\sqrt{1}$$

Simplify.

$$z-9=1 z-9=-1$$

Add 9 to both sides of the equation.

$$z=10 z=8$$