Solve 2*x^2=x+12.3

$$2 \cdot x ^ { 2 } = x + 12.3$$

Solve for x

$$x=\frac{\sqrt{2485}}{20}+0.25\approx 2.742488716$$

$$x=-\frac{\sqrt{2485}}{20}+0.25\approx -2.242488716$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$2x^{2}-x=12.3$$

Subtract 12.3 from both sides.

$$2x^{2}-x-12.3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -12.3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-12.3\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-12.3\right)}}{2\times 2}$$

Multiply -8 times -12.3.

$$x=\frac{-\left(-1\right)±\sqrt{1+98.4}}{2\times 2}$$

Add 1 to 98.4.

$$x=\frac{-\left(-1\right)±\sqrt{99.4}}{2\times 2}$$

Take the square root of 99.4.

$$x=\frac{-\left(-1\right)±\frac{\sqrt{2485}}{5}}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±\frac{\sqrt{2485}}{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±\frac{\sqrt{2485}}{5}}{4}$$

Now solve the equation x=\frac{1±\frac{\sqrt{2485}}{5}}{4} when ± is plus. Add 1 to \frac{\sqrt{2485}}{5}.

$$x=\frac{\frac{\sqrt{2485}}{5}+1}{4}$$

Divide 1+\frac{\sqrt{2485}}{5} by 4.

$$x=\frac{\sqrt{2485}}{20}+\frac{1}{4}$$

Now solve the equation x=\frac{1±\frac{\sqrt{2485}}{5}}{4} when ± is minus. Subtract \frac{\sqrt{2485}}{5} from 1.

$$x=\frac{-\frac{\sqrt{2485}}{5}+1}{4}$$

Divide 1-\frac{\sqrt{2485}}{5} by 4.

$$x=-\frac{\sqrt{2485}}{20}+\frac{1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{2485}}{20}+\frac{1}{4} x=-\frac{\sqrt{2485}}{20}+\frac{1}{4}$$

Steps for Completing the Square

Subtract x from both sides.

$$2x^{2}-x=12.3$$

Divide both sides by 2.

$$\frac{2x^{2}-x}{2}=\frac{12.3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{1}{2}x=\frac{12.3}{2}$$

Divide 12.3 by 2.

$$x^{2}-\frac{1}{2}x=6.15$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=6.15+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=6.15+\frac{1}{16}$$

Add 6.15 to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{497}{80}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{497}{80}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{497}{80}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{2485}}{20} x-\frac{1}{4}=-\frac{\sqrt{2485}}{20}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{\sqrt{2485}}{20}+\frac{1}{4} x=-\frac{\sqrt{2485}}{20}+\frac{1}{4}$$