Solve 0=2c^2-10c-24

$$0 = 2 c ^ { 2 } - 10 c - 24$$

Solve for c

$$c = \frac{\sqrt{73} + 5}{2} \approx 6.772001873$$

$$c=\frac{5-\sqrt{73}}{2}\approx -1.772001873$$

Steps Using the Quadratic Formula

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

$$2c^{2}-10c-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -10 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 2\left(-24\right)}}{2\times 2}$$

Square -10.

$$c=\frac{-\left(-10\right)±\sqrt{100-4\times 2\left(-24\right)}}{2\times 2}$$

Multiply -4 times 2.

$$c=\frac{-\left(-10\right)±\sqrt{100-8\left(-24\right)}}{2\times 2}$$

Multiply -8 times -24.

$$c=\frac{-\left(-10\right)±\sqrt{100+192}}{2\times 2}$$

Add 100 to 192.

$$c=\frac{-\left(-10\right)±\sqrt{292}}{2\times 2}$$

Take the square root of 292.

$$c=\frac{-\left(-10\right)±2\sqrt{73}}{2\times 2}$$

The opposite of -10 is 10.

$$c=\frac{10±2\sqrt{73}}{2\times 2}$$

Multiply 2 times 2.

$$c=\frac{10±2\sqrt{73}}{4}$$

Now solve the equation c=\frac{10±2\sqrt{73}}{4} when ± is plus. Add 10 to 2\sqrt{73}.

$$c=\frac{2\sqrt{73}+10}{4}$$

Divide 10+2\sqrt{73} by 4.

$$c=\frac{\sqrt{73}+5}{2}$$

Now solve the equation c=\frac{10±2\sqrt{73}}{4} when ± is minus. Subtract 2\sqrt{73} from 10.

$$c=\frac{10-2\sqrt{73}}{4}$$

Divide 10-2\sqrt{73} by 4.

$$c=\frac{5-\sqrt{73}}{2}$$

The equation is now solved.

$$c=\frac{\sqrt{73}+5}{2} c=\frac{5-\sqrt{73}}{2}$$

Steps for Completing the Square

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

$$2c^{2}-10c-24=0$$

Add 24 to both sides. Anything plus zero gives itself.

$$2c^{2}-10c=24$$

Divide both sides by 2.

$$\frac{2c^{2}-10c}{2}=\frac{24}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$c^{2}+\left(-\frac{10}{2}\right)c=\frac{24}{2}$$

Divide -10 by 2.

$$c^{2}-5c=\frac{24}{2}$$

Divide 24 by 2.

$$c^{2}-5c=12$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}-5c+\left(-\frac{5}{2}\right)^{2}=12+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$c^{2}-5c+\frac{25}{4}=12+\frac{25}{4}$$

Add 12 to \frac{25}{4}.

$$c^{2}-5c+\frac{25}{4}=\frac{73}{4}$$

Factor c^{2}-5c+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(c-\frac{5}{2}\right)^{2}=\frac{73}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c-\frac{5}{2}\right)^{2}}=\sqrt{\frac{73}{4}}$$

Simplify.

$$c-\frac{5}{2}=\frac{\sqrt{73}}{2} c-\frac{5}{2}=-\frac{\sqrt{73}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$c=\frac{\sqrt{73}+5}{2} c=\frac{5-\sqrt{73}}{2}$$