Solve x+2x^2-26x=2

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

Solve for x

$$x = \frac{\sqrt{641} + 25}{4} \approx 12.579494451$$

$$x=\frac{25-\sqrt{641}}{4}\approx -0.079494451$$

Steps Using the Quadratic Formula

Combine x and -26x to get -25x.

$$-25x+2x^{2}=2$$

Subtract 2 from both sides.

$$-25x+2x^{2}-2=0$$

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.

$$2x^{2}-25x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -25 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 2\left(-2\right)}}{2\times 2}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 2\left(-2\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-25\right)±\sqrt{625-8\left(-2\right)}}{2\times 2}$$

Multiply -8 times -2.

$$x=\frac{-\left(-25\right)±\sqrt{625+16}}{2\times 2}$$

Add 625 to 16.

$$x=\frac{-\left(-25\right)±\sqrt{641}}{2\times 2}$$

The opposite of -25 is 25.

$$x=\frac{25±\sqrt{641}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{25±\sqrt{641}}{4}$$

Now solve the equation x=\frac{25±\sqrt{641}}{4} when ± is plus. Add 25 to \sqrt{641}.

$$x=\frac{\sqrt{641}+25}{4}$$

Now solve the equation x=\frac{25±\sqrt{641}}{4} when ± is minus. Subtract \sqrt{641} from 25.

$$x=\frac{25-\sqrt{641}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{641}+25}{4} x=\frac{25-\sqrt{641}}{4}$$

Steps for Completing the Square

Combine x and -26x to get -25x.

$$-25x+2x^{2}=2$$

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.

$$2x^{2}-25x=2$$

Divide both sides by 2.

$$\frac{2x^{2}-25x}{2}=\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{25}{2}x=\frac{2}{2}$$

Divide 2 by 2.

$$x^{2}-\frac{25}{2}x=1$$

Divide -\frac{25}{2}, the coefficient of the x term, by 2 to get -\frac{25}{4}. Then add the square of -\frac{25}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{25}{2}x+\left(-\frac{25}{4}\right)^{2}=1+\left(-\frac{25}{4}\right)^{2}$$

Square -\frac{25}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{25}{2}x+\frac{625}{16}=1+\frac{625}{16}$$

Add 1 to \frac{625}{16}.

$$x^{2}-\frac{25}{2}x+\frac{625}{16}=\frac{641}{16}$$

Factor x^{2}-\frac{25}{2}x+\frac{625}{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}.

$$\left(x-\frac{25}{4}\right)^{2}=\frac{641}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{4}\right)^{2}}=\sqrt{\frac{641}{16}}$$

Simplify.

$$x-\frac{25}{4}=\frac{\sqrt{641}}{4} x-\frac{25}{4}=-\frac{\sqrt{641}}{4}$$

Add \frac{25}{4} to both sides of the equation.

$$x=\frac{\sqrt{641}+25}{4} x=\frac{25-\sqrt{641}}{4}$$