Solve 59.9=-1x^2+26x+43

$$59.9 = - 1 x ^ { 2 } + 26 x + 43$$

Solve for x

$$x=\frac{39\sqrt{10}}{10}+13\approx 25.332882875$$

$$x=-\frac{39\sqrt{10}}{10}+13\approx 0.667117125$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-x^{2}+26x+43=59.9$$

Subtract 59.9 from both sides.

$$-x^{2}+26x+43-59.9=0$$

Subtract 59.9 from 43 to get -16.9.

$$-x^{2}+26x-16.9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 26 for b, and -16.9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-26±\sqrt{26^{2}-4\left(-1\right)\left(-16.9\right)}}{2\left(-1\right)}$$

Square 26.

$$x=\frac{-26±\sqrt{676-4\left(-1\right)\left(-16.9\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-26±\sqrt{676+4\left(-16.9\right)}}{2\left(-1\right)}$$

Multiply 4 times -16.9.

$$x=\frac{-26±\sqrt{676-67.6}}{2\left(-1\right)}$$

Add 676 to -67.6.

$$x=\frac{-26±\sqrt{608.4}}{2\left(-1\right)}$$

Take the square root of 608.4.

$$x=\frac{-26±\frac{39\sqrt{10}}{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-26±\frac{39\sqrt{10}}{5}}{-2}$$

Now solve the equation x=\frac{-26±\frac{39\sqrt{10}}{5}}{-2} when ± is plus. Add -26 to \frac{39\sqrt{10}}{5}.

$$x=\frac{\frac{39\sqrt{10}}{5}-26}{-2}$$

Divide -26+\frac{39\sqrt{10}}{5} by -2.

$$x=-\frac{39\sqrt{10}}{10}+13$$

Now solve the equation x=\frac{-26±\frac{39\sqrt{10}}{5}}{-2} when ± is minus. Subtract \frac{39\sqrt{10}}{5} from -26.

$$x=\frac{-\frac{39\sqrt{10}}{5}-26}{-2}$$

Divide -26-\frac{39\sqrt{10}}{5} by -2.

$$x=\frac{39\sqrt{10}}{10}+13$$

The equation is now solved.

$$x=-\frac{39\sqrt{10}}{10}+13 x=\frac{39\sqrt{10}}{10}+13$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-x^{2}+26x+43=59.9$$

Subtract 43 from both sides.

$$-x^{2}+26x=59.9-43$$

Subtract 43 from 59.9 to get 16.9.

$$-x^{2}+26x=16.9$$

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.

$$-x^{2}+26x=\frac{169}{10}$$

Divide both sides by -1.

$$\frac{-x^{2}+26x}{-1}=\frac{\frac{169}{10}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{26}{-1}x=\frac{\frac{169}{10}}{-1}$$

Divide 26 by -1.

$$x^{2}-26x=\frac{\frac{169}{10}}{-1}$$

Divide \frac{169}{10} by -1.

$$x^{2}-26x=-\frac{169}{10}$$

Divide -26, the coefficient of the x term, by 2 to get -13. Then add the square of -13 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-26x+\left(-13\right)^{2}=-\frac{169}{10}+\left(-13\right)^{2}$$

Square -13.

$$x^{2}-26x+169=-\frac{169}{10}+169$$

Add -\frac{169}{10} to 169.

$$x^{2}-26x+169=\frac{1521}{10}$$

Factor x^{2}-26x+169. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-13\right)^{2}=\frac{1521}{10}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-13\right)^{2}}=\sqrt{\frac{1521}{10}}$$

Simplify.

$$x-13=\frac{39\sqrt{10}}{10} x-13=-\frac{39\sqrt{10}}{10}$$

Add 13 to both sides of the equation.

$$x=\frac{39\sqrt{10}}{10}+13 x=-\frac{39\sqrt{10}}{10}+13$$