Solve 2x^2+10=12x

$$2 { x }^{ 2 } +10 = 12x$$

Solve for x

$$x=1$$

$$x=5$$

Steps Using Factoring By Grouping

Subtract 12x from both sides.

$$2x^{2}+10-12x=0$$

Divide both sides by 2.

$$x^{2}+5-6x=0$$

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

$$x^{2}-6x+5=0$$

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

$$a+b=-6 ab=1\times 5=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

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

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

Factor out x in the first and -1 in the second group.

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

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

$$\left(x-5\right)\left(x-1\right)$$

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

$$x=5 x=1$$

Steps Using the Quadratic Formula

Subtract 12x from both sides.

$$2x^{2}+10-12x=0$$

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.

$$2x^{2}-12x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -12 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 10}}{2\times 2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 10}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-12\right)±\sqrt{144-8\times 10}}{2\times 2}$$

Multiply -8 times 10.

$$x=\frac{-\left(-12\right)±\sqrt{144-80}}{2\times 2}$$

Add 144 to -80.

$$x=\frac{-\left(-12\right)±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$x=\frac{-\left(-12\right)±8}{2\times 2}$$

The opposite of -12 is 12.

$$x=\frac{12±8}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{12±8}{4}$$

Now solve the equation x=\frac{12±8}{4} when ± is plus. Add 12 to 8.

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

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{12±8}{4} when ± is minus. Subtract 8 from 12.

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

Divide 4 by 4.

$$x=1$$

The equation is now solved.

$$x=5 x=1$$

Steps for Completing the Square

Subtract 12x from both sides.

$$2x^{2}+10-12x=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-12x=-10$$

Divide both sides by 2.

$$\frac{2x^{2}-12x}{2}=-\frac{10}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{12}{2}\right)x=-\frac{10}{2}$$

Divide -12 by 2.

$$x^{2}-6x=-\frac{10}{2}$$

Divide -10 by 2.

$$x^{2}-6x=-5$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-5+9$$

Add -5 to 9.

$$x^{2}-6x+9=4$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-3=2 x-3=-2$$

Add 3 to both sides of the equation.

$$x=5 x=1$$