Solve 2x^2+80=26x

$$2 { x }^{ 2 } +80=26x$$

Solve for x

$$x=5$$

$$x=8$$

Steps Using Factoring By Grouping

Subtract 26x from both sides.

$$2x^{2}+80-26x=0$$

Divide both sides by 2.

$$x^{2}+40-13x=0$$

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

$$x^{2}-13x+40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+40. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 40=40$$

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 40.

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

Calculate the sum for each pair.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

The solution is the pair that gives sum -13.

$$a=-8 b=-5$$

Rewrite x^{2}-13x+40 as \left(x^{2}-8x\right)+\left(-5x+40\right).

$$\left(x^{2}-8x\right)+\left(-5x+40\right)$$

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

$$x\left(x-8\right)-5\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x-5\right)$$

To find equation solutions, solve x-8=0 and x-5=0.

$$x=8 x=5$$

Steps Using the Quadratic Formula

Subtract 26x from both sides.

$$2x^{2}+80-26x=0$$

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.

$$2x^{2}-26x+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -26 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 2\times 80}}{2\times 2}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 2\times 80}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-26\right)±\sqrt{676-8\times 80}}{2\times 2}$$

Multiply -8 times 80.

$$x=\frac{-\left(-26\right)±\sqrt{676-640}}{2\times 2}$$

Add 676 to -640.

$$x=\frac{-\left(-26\right)±\sqrt{36}}{2\times 2}$$

Take the square root of 36.

$$x=\frac{-\left(-26\right)±6}{2\times 2}$$

The opposite of -26 is 26.

$$x=\frac{26±6}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{26±6}{4}$$

Now solve the equation x=\frac{26±6}{4} when ± is plus. Add 26 to 6.

$$x=\frac{32}{4}$$

Divide 32 by 4.

$$x=8$$

Now solve the equation x=\frac{26±6}{4} when ± is minus. Subtract 6 from 26.

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

Divide 20 by 4.

$$x=5$$

The equation is now solved.

$$x=8 x=5$$

Steps for Completing the Square

Subtract 26x from both sides.

$$2x^{2}+80-26x=0$$

Subtract 80 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-26x=-80$$

Divide both sides by 2.

$$\frac{2x^{2}-26x}{2}=-\frac{80}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{26}{2}\right)x=-\frac{80}{2}$$

Divide -26 by 2.

$$x^{2}-13x=-\frac{80}{2}$$

Divide -80 by 2.

$$x^{2}-13x=-40$$

Divide -13, the coefficient of the x term, by 2 to get -\frac{13}{2}. Then add the square of -\frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-13x+\left(-\frac{13}{2}\right)^{2}=-40+\left(-\frac{13}{2}\right)^{2}$$

Square -\frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-13x+\frac{169}{4}=-40+\frac{169}{4}$$

Add -40 to \frac{169}{4}.

$$x^{2}-13x+\frac{169}{4}=\frac{9}{4}$$

Factor x^{2}-13x+\frac{169}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{13}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{13}{2}=\frac{3}{2} x-\frac{13}{2}=-\frac{3}{2}$$

Add \frac{13}{2} to both sides of the equation.

$$x=8 x=5$$