Solve 2x^2-5=15x

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

Solve for x

$$x = \frac{\sqrt{265} + 15}{4} \approx 7.819705149$$

$$x=\frac{15-\sqrt{265}}{4}\approx -0.319705149$$

Steps Using the Quadratic Formula

Subtract 15x from both sides.

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

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.

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

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

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

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-15\right)±\sqrt{225-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$x=\frac{-\left(-15\right)±\sqrt{225+40}}{2\times 2}$$

Add 225 to 40.

$$x=\frac{-\left(-15\right)±\sqrt{265}}{2\times 2}$$

The opposite of -15 is 15.

$$x=\frac{15±\sqrt{265}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{15±\sqrt{265}}{4}$$

Now solve the equation x=\frac{15±\sqrt{265}}{4} when ± is plus. Add 15 to \sqrt{265}.

$$x=\frac{\sqrt{265}+15}{4}$$

Now solve the equation x=\frac{15±\sqrt{265}}{4} when ± is minus. Subtract \sqrt{265} from 15.

$$x=\frac{15-\sqrt{265}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{265}+15}{4} x=\frac{15-\sqrt{265}}{4}$$

Steps for Completing the Square

Subtract 15x from both sides.

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

Add 5 to both sides. Anything plus zero gives itself.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{15}{2}, the coefficient of the x term, by 2 to get -\frac{15}{4}. Then add the square of -\frac{15}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{15}{2}x+\left(-\frac{15}{4}\right)^{2}=\frac{5}{2}+\left(-\frac{15}{4}\right)^{2}$$

Square -\frac{15}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=\frac{5}{2}+\frac{225}{16}$$

Add \frac{5}{2} to \frac{225}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=\frac{265}{16}$$

Factor x^{2}-\frac{15}{2}x+\frac{225}{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}.

$$\left(x-\frac{15}{4}\right)^{2}=\frac{265}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{4}\right)^{2}}=\sqrt{\frac{265}{16}}$$

Simplify.

$$x-\frac{15}{4}=\frac{\sqrt{265}}{4} x-\frac{15}{4}=-\frac{\sqrt{265}}{4}$$

Add \frac{15}{4} to both sides of the equation.

$$x=\frac{\sqrt{265}+15}{4} x=\frac{15-\sqrt{265}}{4}$$