Solve 2x^2-18x+40/2x-10=5

$$\frac { 2 x ^ { 2 } - 18 x + 40 } { 2 x - 10 } = 5$$

Solve for x

$$x=9$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-5\right).

$$2x^{2}-18x+40=10\left(x-5\right)$$

Use the distributive property to multiply 10 by x-5.

$$2x^{2}-18x+40=10x-50$$

Subtract 10x from both sides.

$$2x^{2}-18x+40-10x=-50$$

Combine -18x and -10x to get -28x.

$$2x^{2}-28x+40=-50$$

Add 50 to both sides.

$$2x^{2}-28x+40+50=0$$

Add 40 and 50 to get 90.

$$2x^{2}-28x+90=0$$

Divide both sides by 2.

$$x^{2}-14x+45=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+45. To find a and b, set up a system to be solved.

$$a+b=-14 ab=1\times 45=45$$

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 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -14.

$$a=-9 b=-5$$

Rewrite x^{2}-14x+45 as \left(x^{2}-9x\right)+\left(-5x+45\right).

$$\left(x^{2}-9x\right)+\left(-5x+45\right)$$

Factor out x in the first and -5 in the second group.

$$x\left(x-9\right)-5\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(x-5\right)$$

To find equation solutions, solve x-9=0 and x-5=0.

$$x=9 x=5$$

Variable x cannot be equal to 5.

$$x=9$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-5\right).

$$2x^{2}-18x+40=10\left(x-5\right)$$

Use the distributive property to multiply 10 by x-5.

$$2x^{2}-18x+40=10x-50$$

Subtract 10x from both sides.

$$2x^{2}-18x+40-10x=-50$$

Combine -18x and -10x to get -28x.

$$2x^{2}-28x+40=-50$$

Add 50 to both sides.

$$2x^{2}-28x+40+50=0$$

Add 40 and 50 to get 90.

$$2x^{2}-28x+90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -28 for b, and 90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 2\times 90}}{2\times 2}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 2\times 90}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-28\right)±\sqrt{784-8\times 90}}{2\times 2}$$

Multiply -8 times 90.

$$x=\frac{-\left(-28\right)±\sqrt{784-720}}{2\times 2}$$

Add 784 to -720.

$$x=\frac{-\left(-28\right)±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$x=\frac{-\left(-28\right)±8}{2\times 2}$$

The opposite of -28 is 28.

$$x=\frac{28±8}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{28±8}{4}$$

Now solve the equation x=\frac{28±8}{4} when ± is plus. Add 28 to 8.

$$x=\frac{36}{4}$$

Divide 36 by 4.

$$x=9$$

Now solve the equation x=\frac{28±8}{4} when ± is minus. Subtract 8 from 28.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

The equation is now solved.

$$x=9 x=5$$

Variable x cannot be equal to 5.

$$x=9$$

Steps for Completing the Square

Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-5\right).

$$2x^{2}-18x+40=10\left(x-5\right)$$

Use the distributive property to multiply 10 by x-5.

$$2x^{2}-18x+40=10x-50$$

Subtract 10x from both sides.

$$2x^{2}-18x+40-10x=-50$$

Combine -18x and -10x to get -28x.

$$2x^{2}-28x+40=-50$$

Subtract 40 from both sides.

$$2x^{2}-28x=-50-40$$

Subtract 40 from -50 to get -90.

$$2x^{2}-28x=-90$$

Divide both sides by 2.

$$\frac{2x^{2}-28x}{2}=-\frac{90}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{28}{2}\right)x=-\frac{90}{2}$$

Divide -28 by 2.

$$x^{2}-14x=-\frac{90}{2}$$

Divide -90 by 2.

$$x^{2}-14x=-45$$

Divide -14, the coefficient of the x term, by 2 to get -7. Then add the square of -7 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-14x+\left(-7\right)^{2}=-45+\left(-7\right)^{2}$$

Square -7.

$$x^{2}-14x+49=-45+49$$

Add -45 to 49.

$$x^{2}-14x+49=4$$

Factor x^{2}-14x+49. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-7\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-7\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-7=2 x-7=-2$$

Add 7 to both sides of the equation.

$$x=9 x=5$$

Variable x cannot be equal to 5.

$$x=9$$