Solve 0.002x^2+0.21x-2.3=0

$$0.002 { x }^{ 2 } +0.21x-2.3=0$$

Solve for x

$$x=-115$$

$$x=10$$

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.

$$0.002x^{2}+0.21x-2.3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 0.002 for a, 0.21 for b, and -2.3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-0.21±\sqrt{0.21^{2}-4\times 0.002\left(-2.3\right)}}{2\times 0.002}$$

Square 0.21 by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-0.21±\sqrt{0.0441-4\times 0.002\left(-2.3\right)}}{2\times 0.002}$$

Multiply -4 times 0.002.

$$x=\frac{-0.21±\sqrt{0.0441-0.008\left(-2.3\right)}}{2\times 0.002}$$

Multiply -0.008 times -2.3 by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-0.21±\sqrt{0.0441+0.0184}}{2\times 0.002}$$

Add 0.0441 to 0.0184 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-0.21±\sqrt{0.0625}}{2\times 0.002}$$

Take the square root of 0.0625.

$$x=\frac{-0.21±\frac{1}{4}}{2\times 0.002}$$

Multiply 2 times 0.002.

$$x=\frac{-0.21±\frac{1}{4}}{0.004}$$

Now solve the equation x=\frac{-0.21±\frac{1}{4}}{0.004} when ± is plus. Add -0.21 to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{1}{25}}{0.004}$$

Divide \frac{1}{25} by 0.004 by multiplying \frac{1}{25} by the reciprocal of 0.004.

$$x=10$$

Now solve the equation x=\frac{-0.21±\frac{1}{4}}{0.004} when ± is minus. Subtract \frac{1}{4} from -0.21 by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{23}{50}}{0.004}$$

Divide -\frac{23}{50} by 0.004 by multiplying -\frac{23}{50} by the reciprocal of 0.004.

$$x=-115$$

The equation is now solved.

$$x=10 x=-115$$

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.

$$0.002x^{2}+0.21x-2.3=0$$

Add 2.3 to both sides of the equation.

$$0.002x^{2}+0.21x-2.3-\left(-2.3\right)=-\left(-2.3\right)$$

Subtracting -2.3 from itself leaves 0.

$$0.002x^{2}+0.21x=-\left(-2.3\right)$$

Subtract -2.3 from 0.

$$0.002x^{2}+0.21x=2.3$$

Multiply both sides by 500.

$$\frac{0.002x^{2}+0.21x}{0.002}=\frac{2.3}{0.002}$$

Dividing by 0.002 undoes the multiplication by 0.002.

$$x^{2}+\frac{0.21}{0.002}x=\frac{2.3}{0.002}$$

Divide 0.21 by 0.002 by multiplying 0.21 by the reciprocal of 0.002.

$$x^{2}+105x=\frac{2.3}{0.002}$$

Divide 2.3 by 0.002 by multiplying 2.3 by the reciprocal of 0.002.

$$x^{2}+105x=1150$$

Divide 105, the coefficient of the x term, by 2 to get \frac{105}{2}. Then add the square of \frac{105}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+105x+\left(\frac{105}{2}\right)^{2}=1150+\left(\frac{105}{2}\right)^{2}$$

Square \frac{105}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+105x+\frac{11025}{4}=1150+\frac{11025}{4}$$

Add 1150 to \frac{11025}{4}.

$$x^{2}+105x+\frac{11025}{4}=\frac{15625}{4}$$

Factor x^{2}+105x+\frac{11025}{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}.

$$\left(x+\frac{105}{2}\right)^{2}=\frac{15625}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{105}{2}\right)^{2}}=\sqrt{\frac{15625}{4}}$$

Simplify.

$$x+\frac{105}{2}=\frac{125}{2} x+\frac{105}{2}=-\frac{125}{2}$$

Subtract \frac{105}{2} from both sides of the equation.

$$x=10 x=-115$$