Solve 0.46+a/10+a^2+a/5=1

$$0.46+ \frac{ a }{ 10 } + { a }^{ 2 } + \frac{ a }{ 5 } = 1$$

Solve for a

$$a=-0.9$$

$$a=0.6$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 10, the least common multiple of 10,5.

$$4.6+a+10a^{2}+2a=10$$

Combine a and 2a to get 3a.

$$4.6+3a+10a^{2}=10$$

Subtract 10 from both sides.

$$4.6+3a+10a^{2}-10=0$$

Subtract 10 from 4.6 to get -5.4.

$$-5.4+3a+10a^{2}=0$$

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.

$$10a^{2}+3a-5.4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 3 for b, and -5.4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-3±\sqrt{3^{2}-4\times 10\left(-5.4\right)}}{2\times 10}$$

Square 3.

$$a=\frac{-3±\sqrt{9-4\times 10\left(-5.4\right)}}{2\times 10}$$

Multiply -4 times 10.

$$a=\frac{-3±\sqrt{9-40\left(-5.4\right)}}{2\times 10}$$

Multiply -40 times -5.4.

$$a=\frac{-3±\sqrt{9+216}}{2\times 10}$$

Add 9 to 216.

$$a=\frac{-3±\sqrt{225}}{2\times 10}$$

Take the square root of 225.

$$a=\frac{-3±15}{2\times 10}$$

Multiply 2 times 10.

$$a=\frac{-3±15}{20}$$

Now solve the equation a=\frac{-3±15}{20} when ± is plus. Add -3 to 15.

$$a=\frac{12}{20}$$

Reduce the fraction \frac{12}{20} to lowest terms by extracting and canceling out 4.

$$a=\frac{3}{5}$$

Now solve the equation a=\frac{-3±15}{20} when ± is minus. Subtract 15 from -3.

$$a=-\frac{18}{20}$$

Reduce the fraction \frac{-18}{20} to lowest terms by extracting and canceling out 2.

$$a=-\frac{9}{10}$$

The equation is now solved.

$$a=\frac{3}{5} a=-\frac{9}{10}$$

Steps for Completing the Square

Multiply both sides of the equation by 10, the least common multiple of 10,5.

$$4.6+a+10a^{2}+2a=10$$

Combine a and 2a to get 3a.

$$4.6+3a+10a^{2}=10$$

Subtract 4.6 from both sides.

$$3a+10a^{2}=10-4.6$$

Subtract 4.6 from 10 to get 5.4.

$$3a+10a^{2}=5.4$$

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.

$$10a^{2}+3a=5.4$$

Divide both sides by 10.

$$\frac{10a^{2}+3a}{10}=\frac{5.4}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$a^{2}+\frac{3}{10}a=\frac{5.4}{10}$$

Divide 5.4 by 10.

$$a^{2}+\frac{3}{10}a=0.54$$

Divide \frac{3}{10}, the coefficient of the x term, by 2 to get \frac{3}{20}. Then add the square of \frac{3}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{3}{10}a+\left(\frac{3}{20}\right)^{2}=0.54+\left(\frac{3}{20}\right)^{2}$$

Square \frac{3}{20} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{3}{10}a+\frac{9}{400}=0.54+\frac{9}{400}$$

Add 0.54 to \frac{9}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}+\frac{3}{10}a+\frac{9}{400}=\frac{9}{16}$$

Factor a^{2}+\frac{3}{10}a+\frac{9}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+\frac{3}{20}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{3}{20}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$a+\frac{3}{20}=\frac{3}{4} a+\frac{3}{20}=-\frac{3}{4}$$

Subtract \frac{3}{20} from both sides of the equation.

$$a=\frac{3}{5} a=-\frac{9}{10}$$