Solve 0.3x^2=2x+4

$$0.3 x ^ { 2 } = 2 x + 4$$

Solve for x

$$x = \frac{2 \sqrt{55} + 10}{3} \approx 8.277465658$$

$$x=\frac{10-2\sqrt{55}}{3}\approx -1.610798991$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

$$0.3x^{2}-2x=4$$

Subtract 4 from both sides.

$$0.3x^{2}-2x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 0.3 for a, -2 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 0.3\left(-4\right)}}{2\times 0.3}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 0.3\left(-4\right)}}{2\times 0.3}$$

Multiply -4 times 0.3.

$$x=\frac{-\left(-2\right)±\sqrt{4-1.2\left(-4\right)}}{2\times 0.3}$$

Multiply -1.2 times -4.

$$x=\frac{-\left(-2\right)±\sqrt{4+4.8}}{2\times 0.3}$$

Add 4 to 4.8.

$$x=\frac{-\left(-2\right)±\sqrt{8.8}}{2\times 0.3}$$

Take the square root of 8.8.

$$x=\frac{-\left(-2\right)±\frac{2\sqrt{55}}{5}}{2\times 0.3}$$

The opposite of -2 is 2.

$$x=\frac{2±\frac{2\sqrt{55}}{5}}{2\times 0.3}$$

Multiply 2 times 0.3.

$$x=\frac{2±\frac{2\sqrt{55}}{5}}{0.6}$$

Now solve the equation x=\frac{2±\frac{2\sqrt{55}}{5}}{0.6} when ± is plus. Add 2 to \frac{2\sqrt{55}}{5}.

$$x=\frac{\frac{2\sqrt{55}}{5}+2}{0.6}$$

Divide 2+\frac{2\sqrt{55}}{5} by 0.6 by multiplying 2+\frac{2\sqrt{55}}{5} by the reciprocal of 0.6.

$$x=\frac{2\sqrt{55}+10}{3}$$

Now solve the equation x=\frac{2±\frac{2\sqrt{55}}{5}}{0.6} when ± is minus. Subtract \frac{2\sqrt{55}}{5} from 2.

$$x=\frac{-\frac{2\sqrt{55}}{5}+2}{0.6}$$

Divide 2-\frac{2\sqrt{55}}{5} by 0.6 by multiplying 2-\frac{2\sqrt{55}}{5} by the reciprocal of 0.6.

$$x=\frac{10-2\sqrt{55}}{3}$$

The equation is now solved.

$$x=\frac{2\sqrt{55}+10}{3} x=\frac{10-2\sqrt{55}}{3}$$

Steps for Completing the Square

Subtract 2x from both sides.

$$0.3x^{2}-2x=4$$

Divide both sides of the equation by 0.3, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{0.3x^{2}-2x}{0.3}=\frac{4}{0.3}$$

Dividing by 0.3 undoes the multiplication by 0.3.

$$x^{2}+\left(-\frac{2}{0.3}\right)x=\frac{4}{0.3}$$

Divide -2 by 0.3 by multiplying -2 by the reciprocal of 0.3.

$$x^{2}-\frac{20}{3}x=\frac{4}{0.3}$$

Divide 4 by 0.3 by multiplying 4 by the reciprocal of 0.3.

$$x^{2}-\frac{20}{3}x=\frac{40}{3}$$

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

$$x^{2}-\frac{20}{3}x+\left(-\frac{10}{3}\right)^{2}=\frac{40}{3}+\left(-\frac{10}{3}\right)^{2}$$

Square -\frac{10}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{40}{3}+\frac{100}{9}$$

Add \frac{40}{3} to \frac{100}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{220}{9}$$

Factor x^{2}-\frac{20}{3}x+\frac{100}{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}.

$$\left(x-\frac{10}{3}\right)^{2}=\frac{220}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{10}{3}\right)^{2}}=\sqrt{\frac{220}{9}}$$

Simplify.

$$x-\frac{10}{3}=\frac{2\sqrt{55}}{3} x-\frac{10}{3}=-\frac{2\sqrt{55}}{3}$$

Add \frac{10}{3} to both sides of the equation.

$$x=\frac{2\sqrt{55}+10}{3} x=\frac{10-2\sqrt{55}}{3}$$