Solve 7=15x^2-5x

$$7 = 15 x ^ { 2 } - 5 x$$

Solve for x

$$x=\frac{\sqrt{445}}{30}+\frac{1}{6}\approx 0.869834104$$

$$x=-\frac{\sqrt{445}}{30}+\frac{1}{6}\approx -0.53650077$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$15x^{2}-5x=7$$

Subtract 7 from both sides.

$$15x^{2}-5x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -5 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 15\left(-7\right)}}{2\times 15}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 15\left(-7\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-5\right)±\sqrt{25-60\left(-7\right)}}{2\times 15}$$

Multiply -60 times -7.

$$x=\frac{-\left(-5\right)±\sqrt{25+420}}{2\times 15}$$

Add 25 to 420.

$$x=\frac{-\left(-5\right)±\sqrt{445}}{2\times 15}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{445}}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{5±\sqrt{445}}{30}$$

Now solve the equation x=\frac{5±\sqrt{445}}{30} when ± is plus. Add 5 to \sqrt{445}.

$$x=\frac{\sqrt{445}+5}{30}$$

Divide 5+\sqrt{445} by 30.

$$x=\frac{\sqrt{445}}{30}+\frac{1}{6}$$

Now solve the equation x=\frac{5±\sqrt{445}}{30} when ± is minus. Subtract \sqrt{445} from 5.

$$x=\frac{5-\sqrt{445}}{30}$$

Divide 5-\sqrt{445} by 30.

$$x=-\frac{\sqrt{445}}{30}+\frac{1}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{445}}{30}+\frac{1}{6} x=-\frac{\sqrt{445}}{30}+\frac{1}{6}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$15x^{2}-5x=7$$

Divide both sides by 15.

$$\frac{15x^{2}-5x}{15}=\frac{7}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$x^{2}+\left(-\frac{5}{15}\right)x=\frac{7}{15}$$

Reduce the fraction \frac{-5}{15} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{1}{3}x=\frac{7}{15}$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{7}{15}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{7}{15}+\frac{1}{36}$$

Add \frac{7}{15} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{89}{180}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{89}{180}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{89}{180}}$$

Simplify.

$$x-\frac{1}{6}=\frac{\sqrt{445}}{30} x-\frac{1}{6}=-\frac{\sqrt{445}}{30}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{\sqrt{445}}{30}+\frac{1}{6} x=-\frac{\sqrt{445}}{30}+\frac{1}{6}$$