Solve -0.1x^2+11x=280

$$- 0.1 x ^ { 2 } + 11 x = 280$$

Solve for x

$$x=40$$

$$x=70$$

Steps Using the Quadratic Formula

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.

$$-0.1x^{2}+11x=280$$

Subtract 280 from both sides of the equation.

$$-0.1x^{2}+11x-280=280-280$$

Subtracting 280 from itself leaves 0.

$$-0.1x^{2}+11x-280=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -0.1 for a, 11 for b, and -280 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-0.1\right)\left(-280\right)}}{2\left(-0.1\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-0.1\right)\left(-280\right)}}{2\left(-0.1\right)}$$

Multiply -4 times -0.1.

$$x=\frac{-11±\sqrt{121+0.4\left(-280\right)}}{2\left(-0.1\right)}$$

Multiply 0.4 times -280.

$$x=\frac{-11±\sqrt{121-112}}{2\left(-0.1\right)}$$

Add 121 to -112.

$$x=\frac{-11±\sqrt{9}}{2\left(-0.1\right)}$$

Take the square root of 9.

$$x=\frac{-11±3}{2\left(-0.1\right)}$$

Multiply 2 times -0.1.

$$x=\frac{-11±3}{-0.2}$$

Now solve the equation x=\frac{-11±3}{-0.2} when ± is plus. Add -11 to 3.

$$x=-\frac{8}{-0.2}$$

Divide -8 by -0.2 by multiplying -8 by the reciprocal of -0.2.

$$x=40$$

Now solve the equation x=\frac{-11±3}{-0.2} when ± is minus. Subtract 3 from -11.

$$x=-\frac{14}{-0.2}$$

Divide -14 by -0.2 by multiplying -14 by the reciprocal of -0.2.

$$x=70$$

The equation is now solved.

$$x=40 x=70$$

Steps for Completing the Square

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.

$$-0.1x^{2}+11x=280$$

Multiply both sides by -10.

$$\frac{-0.1x^{2}+11x}{-0.1}=\frac{280}{-0.1}$$

Dividing by -0.1 undoes the multiplication by -0.1.

$$x^{2}+\frac{11}{-0.1}x=\frac{280}{-0.1}$$

Divide 11 by -0.1 by multiplying 11 by the reciprocal of -0.1.

$$x^{2}-110x=\frac{280}{-0.1}$$

Divide 280 by -0.1 by multiplying 280 by the reciprocal of -0.1.

$$x^{2}-110x=-2800$$

Divide -110, the coefficient of the x term, by 2 to get -55. Then add the square of -55 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-110x+\left(-55\right)^{2}=-2800+\left(-55\right)^{2}$$

Square -55.

$$x^{2}-110x+3025=-2800+3025$$

Add -2800 to 3025.

$$x^{2}-110x+3025=225$$

Factor x^{2}-110x+3025. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-55\right)^{2}=225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-55\right)^{2}}=\sqrt{225}$$

Simplify.

$$x-55=15 x-55=-15$$

Add 55 to both sides of the equation.

$$x=70 x=40$$