Solve (25-2x+1)x=80

$$(25-2x+1)x=80$$

Solve for x

$$x=5$$

$$x=8$$

Steps Using the Quadratic Formula

Add 25 and 1 to get 26.

$$\left(26-2x\right)x=80$$

Use the distributive property to multiply 26-2x by x.

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

Subtract 80 from both sides.

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

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}.

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

Square 26.

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

Multiply -4 times -2.

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

Multiply 8 times -80.

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

Add 676 to -640.

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

Take the square root of 36.

$$x=\frac{-26±6}{2\left(-2\right)}$$

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.

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

Divide -20 by -4.

$$x=5$$

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

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

Divide -32 by -4.

$$x=8$$

The equation is now solved.

$$x=5 x=8$$

Steps for Completing the Square

Add 25 and 1 to get 26.

$$\left(26-2x\right)x=80$$

Use the distributive property to multiply 26-2x by x.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

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.

$$x=8 x=5$$