Solve 4a=-2+3a^2

$$4 a = - 2 + 3 a ^ { 2 }$$

Solve for a

$$a = \frac{\sqrt{10} + 2}{3} \approx 1.72075922$$

$$a=\frac{2-\sqrt{10}}{3}\approx -0.387425887$$

Steps Using the Quadratic Formula

Subtract -2 from both sides.

$$4a-\left(-2\right)=3a^{2}$$

The opposite of -2 is 2.

$$4a+2=3a^{2}$$

Subtract 3a^{2} from both sides.

$$4a+2-3a^{2}=0$$

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.

$$-3a^{2}+4a+2=0$$

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

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

Square 4.

$$a=\frac{-4±\sqrt{16-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$a=\frac{-4±\sqrt{16+12\times 2}}{2\left(-3\right)}$$

Multiply 12 times 2.

$$a=\frac{-4±\sqrt{16+24}}{2\left(-3\right)}$$

Add 16 to 24.

$$a=\frac{-4±\sqrt{40}}{2\left(-3\right)}$$

Take the square root of 40.

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

Multiply 2 times -3.

$$a=\frac{-4±2\sqrt{10}}{-6}$$

Now solve the equation a=\frac{-4±2\sqrt{10}}{-6} when ± is plus. Add -4 to 2\sqrt{10}.

$$a=\frac{2\sqrt{10}-4}{-6}$$

Divide -4+2\sqrt{10} by -6.

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

Now solve the equation a=\frac{-4±2\sqrt{10}}{-6} when ± is minus. Subtract 2\sqrt{10} from -4.

$$a=\frac{-2\sqrt{10}-4}{-6}$$

Divide -4-2\sqrt{10} by -6.

$$a=\frac{\sqrt{10}+2}{3}$$

The equation is now solved.

$$a=\frac{2-\sqrt{10}}{3} a=\frac{\sqrt{10}+2}{3}$$

Steps for Completing the Square

Subtract 3a^{2} from both sides.

$$4a-3a^{2}=-2$$

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.

$$-3a^{2}+4a=-2$$

Divide both sides by -3.

$$\frac{-3a^{2}+4a}{-3}=-\frac{2}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$a^{2}+\frac{4}{-3}a=-\frac{2}{-3}$$

Divide 4 by -3.

$$a^{2}-\frac{4}{3}a=-\frac{2}{-3}$$

Divide -2 by -3.

$$a^{2}-\frac{4}{3}a=\frac{2}{3}$$

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{4}{3}a+\left(-\frac{2}{3}\right)^{2}=\frac{2}{3}+\left(-\frac{2}{3}\right)^{2}$$

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{4}{3}a+\frac{4}{9}=\frac{2}{3}+\frac{4}{9}$$

Add \frac{2}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{4}{3}a+\frac{4}{9}=\frac{10}{9}$$

Factor a^{2}-\frac{4}{3}a+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{2}{3}\right)^{2}=\frac{10}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{2}{3}\right)^{2}}=\sqrt{\frac{10}{9}}$$

Simplify.

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

Add \frac{2}{3} to both sides of the equation.

$$a=\frac{\sqrt{10}+2}{3} a=\frac{2-\sqrt{10}}{3}$$