Solve 2x^2=16x-30

$$2 x ^ { 2 } = 16 x - 30$$

Solve for x

$$x=3$$

$$x=5$$

Steps Using Factoring By Grouping

Subtract 16x from both sides.

$$2x^{2}-16x=-30$$

Add 30 to both sides.

$$2x^{2}-16x+30=0$$

Divide both sides by 2.

$$x^{2}-8x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 15=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

Rewrite x^{2}-8x+15 as \left(x^{2}-5x\right)+\left(-3x+15\right).

$$\left(x^{2}-5x\right)+\left(-3x+15\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-5\right)-3\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x-3\right)$$

To find equation solutions, solve x-5=0 and x-3=0.

$$x=5 x=3$$

Steps Using the Quadratic Formula

Subtract 16x from both sides.

$$2x^{2}-16x=-30$$

Add 30 to both sides.

$$2x^{2}-16x+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -16 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 2\times 30}}{2\times 2}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 2\times 30}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-16\right)±\sqrt{256-8\times 30}}{2\times 2}$$

Multiply -8 times 30.

$$x=\frac{-\left(-16\right)±\sqrt{256-240}}{2\times 2}$$

Add 256 to -240.

$$x=\frac{-\left(-16\right)±\sqrt{16}}{2\times 2}$$

Take the square root of 16.

$$x=\frac{-\left(-16\right)±4}{2\times 2}$$

The opposite of -16 is 16.

$$x=\frac{16±4}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{16±4}{4}$$

Now solve the equation x=\frac{16±4}{4} when ± is plus. Add 16 to 4.

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

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{16±4}{4} when ± is minus. Subtract 4 from 16.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

The equation is now solved.

$$x=5 x=3$$

Steps for Completing the Square

Subtract 16x from both sides.

$$2x^{2}-16x=-30$$

Divide both sides by 2.

$$\frac{2x^{2}-16x}{2}=-\frac{30}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{16}{2}\right)x=-\frac{30}{2}$$

Divide -16 by 2.

$$x^{2}-8x=-\frac{30}{2}$$

Divide -30 by 2.

$$x^{2}-8x=-15$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-15+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-15+16$$

Add -15 to 16.

$$x^{2}-8x+16=1$$

Factor x^{2}-8x+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}.

$$\left(x-4\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-4=1 x-4=-1$$

Add 4 to both sides of the equation.

$$x=5 x=3$$